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Nilai yang memenuhi pertidaksamaan:  adalah ...

Pertanyaan

Nilai x yang memenuhi pertidaksamaan: scriptbase log invisible function application left parenthesis 2 x plus 4 right parenthesis end scriptbase presuperscript 2 less than 3 adalah ...

  1. x less than 2

  2. x less than negative 2

  3. 0 less than x less than 2

  4. negative 2 less than x less than 2

  5. negative 2 less than x less than 0

Pembahasan Soal:

Ingat pada pertidaksamaan bentuk logaritma jika scriptbase log invisible function application f left parenthesis x right parenthesis end scriptbase presuperscript a less than scriptbase log invisible function application b end scriptbase presuperscript a dan a greater than 0 maka 0 less than f left parenthesis x right parenthesis less than b.

Ingat juga sifat-sifat pada bentuk bentuk logaritma yaitu  

  • scriptbase log invisible function application a end scriptbase presuperscript a equals 1
  • scriptbase log invisible function application b to the power of m end scriptbase presuperscript a equals m times scriptbase log invisible function application b end scriptbase presuperscript a

Sehingga akan diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell scriptbase log invisible function application open parentheses 2 x plus 4 close parentheses end scriptbase presuperscript 2 end cell less than 3 row cell scriptbase log invisible function application left parenthesis 2 x plus 4 right parenthesis end scriptbase presuperscript 2 end cell less than cell 3 times scriptbase log invisible function application 2 end scriptbase presuperscript 2 end cell row cell scriptbase log invisible function application left parenthesis 2 x plus 4 right parenthesis end scriptbase presuperscript 2 end cell less than cell scriptbase log invisible function application 2 cubed end scriptbase presuperscript 2 end cell row cell scriptbase log invisible function application left parenthesis 2 x plus 4 right parenthesis end scriptbase presuperscript 2 end cell less than cell scriptbase log invisible function application 8 end scriptbase presuperscript 2 end cell row cell 2 x plus 4 end cell less than 8 row cell 2 x end cell less than cell 8 minus 4 end cell row cell 2 x end cell less than 4 row x less than 2 end table

Karena syarat numerus pada bentuk logaritma scriptbase log invisible function application b end scriptbase presuperscript a yaitu b greater than 0, maka

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x plus 4 end cell greater than 0 row cell 2 x end cell greater than cell negative 4 end cell row x greater than cell fraction numerator negative 4 over denominator 2 end fraction end cell row x greater than cell negative 2 end cell end table

Dari hasil di atas maka penyelesainnya yaitu irisan dari x less than 2 dan x greater than negative 2 yaitu negative 2 less than x less than 2.

Jadi, dapat disimpulkan bahwa nilai x yang memenuhi pertidaksamaan: scriptbase log invisible function application left parenthesis 2 x plus 4 right parenthesis end scriptbase presuperscript 2 less than 3 adalah negative 2 less than x less than 2.

Oleh karena itu, jawaban yang benar adalah D.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

M. Iqbal

Mahasiswa/Alumni Universitas Negeri Semarang

Terakhir diupdate 11 Juli 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Penyelesaian dari pertidaksamaan:  adalah ...

Pembahasan Soal:

Ingat pada pertidaksamaan bentuk logaritma jika scriptbase log invisible function application f left parenthesis x right parenthesis end scriptbase presuperscript a less than scriptbase log invisible function application b end scriptbase presuperscript a dan 0 less than a less than 1 maka f open parentheses x close parentheses greater than b greater than 0.

Ingat juga sifat-sifat pada bentuk bentuk logaritma yaitu  

  • scriptbase log invisible function application a end scriptbase presuperscript a equals 1
  • scriptbase log invisible function application b to the power of m end scriptbase presuperscript a equals m times scriptbase log invisible function application b end scriptbase presuperscript a
  • scriptbase log invisible function application b to the power of m end scriptbase presuperscript a to the power of n end presuperscript equals m over n times scriptbase log invisible function application b end scriptbase presuperscript a

Sehingga akan diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell scriptbase log invisible function application open parentheses 2 x plus 5 close parentheses end scriptbase presuperscript 1 third end presuperscript end cell less than cell negative 2 end cell row cell scriptbase log invisible function application left parenthesis 2 x plus 5 right parenthesis end scriptbase presuperscript 3 to the power of negative 1 end exponent end presuperscript end cell less than cell negative 2 times scriptbase log invisible function application 3 end scriptbase presuperscript 3 end cell row cell negative 1 times scriptbase log invisible function application left parenthesis 2 x plus 5 right parenthesis end scriptbase presuperscript 3 end cell less than cell negative 2 times scriptbase log invisible function application 3 end scriptbase presuperscript 3 end cell row cell scriptbase log invisible function application left parenthesis 2 x plus 5 right parenthesis to the power of negative 1 end exponent end scriptbase presuperscript 3 end cell less than cell scriptbase log invisible function application open parentheses 3 close parentheses to the power of negative 2 end exponent end scriptbase presuperscript 3 end cell row cell left parenthesis 2 x plus 5 right parenthesis to the power of negative 1 end exponent end cell less than cell open parentheses 3 close parentheses to the power of negative 2 end exponent end cell row cell left parenthesis 2 x plus 5 right parenthesis to the power of 1 end cell greater than cell open parentheses 3 close parentheses squared end cell row cell 2 x plus 5 end cell greater than 9 row cell 2 x end cell greater than cell 9 minus 5 end cell row cell 2 x end cell greater than 4 row x greater than cell 4 over 2 end cell row x greater than 2 end table

Karena syarat numerus pada bentuk logaritma scriptbase log invisible function application b end scriptbase presuperscript a yaitu b greater than 0, maka

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x plus 5 end cell greater than 0 row cell 2 x end cell greater than cell negative 5 end cell row x greater than cell fraction numerator negative 5 over denominator 2 end fraction end cell row x greater than cell negative 2 , 5 end cell end table

Dari hasil di atas maka penyelesainnya yaitu irisan dari x greater than 2 dan x greater than negative 2 , 5 yaitu x greater than 2

Jadi, dapat disimpulkan bahwa penyelesaian dari pertidaksamaan scriptbase log invisible function application left parenthesis 2 x plus 5 right parenthesis end scriptbase presuperscript 1 third end presuperscript less than negative 2 adalah x greater than 2.

Oleh karena itu, jawaban yang benar adalah A.

0

Roboguru

Nilai yang memenuhi pertidaksamaan:  adalah ...

Pembahasan Soal:

Ingat pada pertidaksamaan bentuk logaritma jika scriptbase log invisible function application f left parenthesis x right parenthesis end scriptbase presuperscript a greater than scriptbase log invisible function application b end scriptbase presuperscript a dan 0 less than a less than 1 maka 0 less than f open parentheses x close parentheses less than b.

Ingat juga sifat-sifat pada bentuk bentuk logaritma yaitu  

  • scriptbase log invisible function application a end scriptbase presuperscript a equals 1
  • scriptbase log invisible function application b to the power of m end scriptbase presuperscript a equals m times scriptbase log invisible function application b end scriptbase presuperscript a
  • scriptbase log invisible function application b to the power of m end scriptbase presuperscript a to the power of n end presuperscript equals m over n times scriptbase log invisible function application b end scriptbase presuperscript a

Sehingga akan diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell scriptbase log invisible function application open parentheses x squared minus 1 close parentheses end scriptbase presuperscript 0 , 1 end presuperscript end cell greater than 2 row cell scriptbase log invisible function application left parenthesis x squared minus 1 right parenthesis end scriptbase presuperscript 1 over 10 end presuperscript end cell greater than cell 2 times scriptbase log invisible function application 10 end scriptbase presuperscript 10 end cell row cell scriptbase log invisible function application left parenthesis x squared minus 1 right parenthesis end scriptbase presuperscript 10 to the power of negative 1 end exponent end presuperscript end cell greater than cell scriptbase log invisible function application 10 squared end scriptbase presuperscript 10 end cell row cell negative 1 times scriptbase log invisible function application left parenthesis x squared minus 1 right parenthesis end scriptbase presuperscript 10 end cell greater than cell scriptbase log invisible function application 100 end scriptbase presuperscript 10 end cell row cell scriptbase log invisible function application left parenthesis x squared minus 1 right parenthesis to the power of negative 1 end exponent end scriptbase presuperscript 10 end cell greater than cell scriptbase log invisible function application 100 end scriptbase presuperscript 10 end cell row cell left parenthesis x squared minus 1 right parenthesis to the power of negative 1 end exponent end cell greater than 100 row cell left parenthesis x squared minus 1 right parenthesis to the power of 1 end cell less than cell 100 to the power of negative 1 end exponent end cell row cell x squared minus 1 end cell less than cell 1 over 100 end cell row cell x squared minus 1 end cell less than cell 0 , 01 end cell row cell x squared end cell less than cell 0 , 01 plus 1 end cell row cell x squared end cell less than cell 1 , 01 end cell row x less than cell square root of 1 , 01 end root end cell row x less than cell plus-or-minus square root of 1 , 01 end root end cell end table

Didapatkan x less than square root of 1 , 01 end root atau x greater than negative square root of 1 , 01 end root.

Karena syarat numerus pada bentuk logaritma scriptbase log invisible function application b end scriptbase presuperscript a yaitu b greater than 0, maka

table attributes columnalign right center left columnspacing 0px end attributes row cell x squared minus 1 end cell greater than 0 row cell x squared end cell greater than 1 row x greater than cell square root of 1 end cell row x greater than cell plus-or-minus 1 end cell end table

Didapatkan x greater than 1 dan x less than negative 1

Dari hasil di atas maka penyelesainnya yaitu irisan dari x less than square root of 1 , 01 end rootx greater than negative square root of 1 , 01 end rootx greater than 1, dan x less than negative 1 yaitu negative square root of 1 , 01 end root less than x less than negative 1 atau 1 less than x less than square root of 1 , 01 end root.

Jadi, dapat disimpulkan bahwa nilai x yang memenuhi pertidaksamaan: scriptbase log invisible function application left parenthesis x squared minus 1 right parenthesis end scriptbase presuperscript 0 , 1 end presuperscript greater than 2 adalah negative square root of 1 , 01 end root less than x less than negative 1 atau 1 less than x less than square root of 1 , 01 end root.

Oleh karena itu, jawaban yang benar adalah D.

0

Roboguru

Himpunan penyelesaian pertidaksamaan  adalah ...

Pembahasan Soal:

Ingat sifat bentuk logaritma:

table attributes columnalign right center left columnspacing 0px end attributes row cell log presuperscript a space b plus log presuperscript a space c end cell equals cell log presuperscript a space b c end cell row cell log presuperscript a space a to the power of n end cell equals n end table 

dan ingat pada pertidaksamaan logaritma berlaku:

log presuperscript a space f open parentheses x close parentheses less or equal than log presuperscript a space g open parentheses x close parentheses rightwards arrow f open parentheses x close parentheses less or equal than g open parentheses x close parentheses comma space f open parentheses x close parentheses greater than 0 comma space g open parentheses x close parentheses greater than 0 

Sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell log presuperscript 2 space open parentheses x minus 2 close parentheses plus log presuperscript 2 space open parentheses x plus 5 close parentheses end cell less or equal than 3 row cell log presuperscript 2 space open parentheses x minus 2 close parentheses open parentheses x plus 5 close parentheses end cell less or equal than cell log presuperscript 2 space 2 cubed end cell row cell log presuperscript 2 space open parentheses x squared plus 3 x minus 10 close parentheses end cell less or equal than cell log presuperscript 2 space 2 cubed end cell row cell x squared plus 3 x minus 10 end cell less or equal than cell 2 cubed end cell row cell x squared plus 3 x minus 10 end cell less or equal than 8 row cell x squared plus 3 x minus 10 minus 8 end cell less or equal than 0 row cell x squared plus 3 x minus 18 end cell less or equal than 0 row cell open parentheses x plus 6 close parentheses open parentheses x minus 3 close parentheses end cell less or equal than 0 end table 

sehingga nilai x yang memenuhi adalah negative 6 less or equal than x less or equal than 3.

Syarat numerus:

table row cell x minus 2 greater than 0 end cell rightwards arrow cell x greater than 2 end cell row cell x plus 5 greater than 0 end cell rightwards arrow cell x greater than negative 5 end cell end table 

Diperoleh x greater than negative 5x greater than 2 dan negative 6 less or equal than x less or equal than 3 maka irisannya adalah 2 less than x less or equal than 3. Dengan demikian himpunan penyelesaian dari pertidaksamaan tersebut adalah open curly brackets x vertical line 2 less than x less or equal than 3 comma space x element of straight R close curly brackets.

Oleh karena itu, jawaban yang benar adalah E.

 

0

Roboguru

Himpunan penyelesaian pertidaksamaan  adalah. . . .

Pembahasan Soal:

Diketahui pertidaksamaan log presuperscript 2 open parentheses x minus 2 close parentheses plus log presuperscript 2 open parentheses x plus 5 close parentheses less or equal than 3. Dengan menggunakan sifat bentuk logaritma berikut.

untuk a comma space b comma space c greater than 0 dan a not equal to 1, berlaku:

  1. log presuperscript a space open parentheses b c close parentheses equals log presuperscript a space b plus log presuperscript a space c
  2.  log presuperscript a space a equals 1
  3. log presuperscript a space b to the power of m equals m cross times log presuperscript a space b

maka diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell log presuperscript 2 open parentheses x minus 2 close parentheses plus log presuperscript 2 open parentheses x plus 5 close parentheses end cell less or equal than 3 row cell log presuperscript 2 open parentheses x minus 2 close parentheses open parentheses x plus 5 close parentheses end cell less or equal than cell 3 cross times 1 end cell row cell log presuperscript 2 open parentheses x minus 2 close parentheses open parentheses x plus 5 close parentheses end cell less or equal than cell 3 cross times log presuperscript 2 space 2 end cell row cell log presuperscript 2 open parentheses x minus 2 close parentheses open parentheses x plus 5 close parentheses end cell less or equal than cell log presuperscript 2 space 2 cubed end cell row cell log presuperscript 2 open parentheses x minus 2 close parentheses open parentheses x plus 5 close parentheses end cell less or equal than cell log presuperscript 2 space 8 end cell end table

Kemudian, ingat bahwa, jika log presuperscript a space f open parentheses x close parentheses less or equal than log presuperscript a space pa greater than 1 dan p greater than 0 maka f open parentheses x close parentheses less or equal than p dan f open parentheses x close parentheses greater than 0.

Misal, f open parentheses x close parentheses equals open parentheses x minus 2 close parentheses open parentheses x plus 5 close parentheses maka dari pertidaksamaan log presuperscript 2 open parentheses x minus 2 close parentheses open parentheses x plus 5 close parentheses less or equal than log presuperscript 2 space 8 diperoleh:

  • f open parentheses x close parentheses less or equal than 8

 table attributes columnalign right center left columnspacing 0px end attributes row cell f open parentheses x close parentheses end cell less or equal than 8 row cell open parentheses x minus 2 close parentheses open parentheses x plus 5 close parentheses end cell less or equal than 8 row cell x squared plus 3 x minus 10 end cell less or equal than 8 row cell x squared plus 3 x minus 10 minus 8 end cell less or equal than 0 row cell x squared plus 3 x minus 18 end cell less or equal than 0 row cell open parentheses x plus 6 close parentheses open parentheses x minus 3 close parentheses end cell less or equal than 0 end table

Nilai x yang memenuhi saat sama dengan 0 sebagai berikut.

open parentheses x plus 6 close parentheses open parentheses x minus 3 close parentheses equals 0 table row cell x equals negative 6 end cell atau cell x equals 3 end cell end table

Lalu, nilai x di atas dapat dituliskan pada garis bilangan sebagai berikut.

Berdasarkan garis bilangan di atas, terdapat 3 daerah yaitu x less than negative 6negative 6 less than x less than 3, dan x greater than 3. Uji titik pada setiap daerah tersebut sebagai berikut.

Ketika x less than negative 6, pilih x equals negative 7 maka diperoleh:

open parentheses x plus 6 close parentheses open parentheses x minus 3 close parentheses rightwards double arrowtable attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses negative 7 plus 6 close parentheses times open parentheses negative 7 minus 3 close parentheses end cell equals cell open parentheses negative 1 close parentheses times open parentheses negative 10 close parentheses equals 10 space greater than 0 end cell end table

Berdasarkan uji titik di atas, ketika x less than negative 6, nilai open parentheses x plus 6 close parentheses open parentheses x minus 3 close parentheses lebih dari 0.

Ketika negative 6 less than x less than 3, pilih x equals 0, maka diperoleh:

open parentheses x plus 6 close parentheses open parentheses x minus 3 close parentheses rightwards double arrowtable attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 0 plus 6 close parentheses times open parentheses 0 minus 3 close parentheses end cell equals cell 6 times open parentheses negative 3 close parentheses equals negative 18 space less than 0 end cell end table

Berdasarkan uji titik di atas, ketika negative 6 less than x less than 3, nilai open parentheses x plus 6 close parentheses open parentheses x minus 3 close parentheses kurang dari 0.

Ketika x greater than 3, pilih x equals 4, maka diperoleh:

open parentheses x plus 6 close parentheses open parentheses x minus 3 close parentheses rightwards double arrowtable attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 4 plus 6 close parentheses times open parentheses 4 minus 3 close parentheses end cell equals cell 10 times 1 equals 10 space greater than 0 end cell end table

Berdasarkan uji titik di atas, ketika x greater than 3, nilai open parentheses x plus 6 close parentheses open parentheses x minus 3 close parentheses lebih dari 0.

Hasil di atas dapat dituliskan pada garis bilangan sebagai berikut.

Berdasarkan garis bilangan di atas, nilai x yang memenuhi pertidaksamaan open parentheses x plus 6 close parentheses open parentheses x minus 3 close parentheses less or equal than 0 adalah negative 6 less than x less than 3.

  • f open parentheses x close parentheses greater than 0

table attributes columnalign right center left columnspacing 0px end attributes row cell f open parentheses x close parentheses end cell greater than 0 row cell open parentheses x minus 2 close parentheses open parentheses x plus 5 close parentheses end cell greater than 0 end table

Nilai x yang memenuhi saat sama dengan 0 sebagai berikut.

open parentheses x minus 2 close parentheses open parentheses x plus 5 close parentheses equals 0 table row cell x equals 2 end cell atau cell x equals negative 5 end cell end table

Lalu, nilai x di atas dapat dituliskan pada garis bilangan sebagai berikut.

Berdasarkan garis bilangan di atas, terdapat 3 daerah yaitu x less than negative 5negative 5 less than x less than 2, dan x greater than 2. Uji titik pada setiap daerah tersebut sebagai berikut.

Ketika x less than negative 5, pilih x equals negative 6 maka diperoleh:

open parentheses x minus 2 close parentheses open parentheses x plus 5 close parentheses rightwards double arrowtable attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses negative 6 minus 2 close parentheses times open parentheses negative 6 plus 5 close parentheses end cell equals cell open parentheses negative 8 close parentheses times open parentheses negative 1 close parentheses equals 8 space greater than 0 end cell end table

Berdasarkan uji titik di atas, ketika x less than negative 5, nilai open parentheses x minus 2 close parentheses open parentheses x plus 5 close parentheses lebih dari 0.

Ketika negative 5 less than x less than 2, pilih x equals 0, maka diperoleh:

open parentheses x minus 2 close parentheses open parentheses x plus 5 close parentheses rightwards double arrowtable attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 0 minus 2 close parentheses times open parentheses 0 plus 5 close parentheses end cell equals cell open parentheses negative 2 close parentheses times 5 equals negative 10 space less than 0 end cell end table

Berdasarkan uji titik di atas, ketika negative 5 less than x less than 2, nilai open parentheses x minus 2 close parentheses open parentheses x plus 5 close parentheses kurang dari 0.

Ketika x greater than 2, pilih x equals 3, maka diperoleh:

open parentheses x minus 2 close parentheses open parentheses x plus 5 close parentheses rightwards double arrowtable attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 3 minus 2 close parentheses times open parentheses 3 plus 5 close parentheses end cell equals cell 1 times 8 equals 8 space greater than 0 end cell end table

Berdasarkan uji titik di atas, ketika x greater than 2, nilai open parentheses x minus 2 close parentheses open parentheses x plus 5 close parentheses lebih dari 0.

Hasil di atas dapat dituliskan pada garis bilangan sebagai berikut.

Berdasarkan garis bilangan di atas, nilai x yang memenuhi pertidaksamaan open parentheses x minus 2 close parentheses open parentheses x plus 5 close parentheses greater than 0 adalah x less than negative 5 atau x greater than 2.

Kemudian, numerus dari suatu bentuk logaritma bernilai lebih dari 0, maka dari log presuperscript 2 open parentheses x minus 2 close parentheses plus log presuperscript 2 open parentheses x plus 5 close parentheses less or equal than 3 terdapat syarat x minus 2 greater than 0 dan x plus 5 greater than 0. Jika x minus 2 greater than 0, maka x greater than 2 dan jika x plus 5 greater than 0, maka x greater than negative 5.

Lalu, penyelesaian dari log presuperscript 2 open parentheses x minus 2 close parentheses plus log presuperscript 2 open parentheses x plus 5 close parentheses less or equal than 3 adalah irisan dari penyelesaian f open parentheses x close parentheses less or equal than 8f open parentheses x close parentheses greater than 0x minus 2 greater than 0, dan x plus 5 greater than 0. Irisan dari keempat penyelesaian tersebut dapat ditentukan dengan garis bilangan sebagai berikut.

 

Berdasarkan garis bilangan di atas, himpunan penyelesaian pertidaksamaan log presuperscript 2 open parentheses x minus 2 close parentheses plus log presuperscript 2 open parentheses x plus 5 close parentheses less or equal than 3 adalah open curly brackets x vertical line space 2 less than x less or equal than 3 close curly brackets.

Oleh karena itu, jawaban yang benar adalah D.

0

Roboguru

Jika , nilai x yang memenuhi adalah...

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row cell log presuperscript 2 space open parentheses x squared plus x plus 4 close parentheses end cell less than cell log presuperscript 5 space 625 end cell row cell log presuperscript 2 space open parentheses x squared plus x plus 4 close parentheses end cell less than cell log presuperscript 5 space 5 to the power of 4 end cell row cell log presuperscript 2 space open parentheses x squared plus x plus 4 close parentheses end cell less than cell 4 cross times log presuperscript 5 5 end cell row cell log presuperscript 2 space open parentheses x squared plus x plus 4 close parentheses end cell less than cell 4 cross times 1 end cell row cell log presuperscript 2 space open parentheses x squared plus x plus 4 close parentheses end cell less than 4 row cell log presuperscript 2 space open parentheses x squared plus x plus 4 close parentheses end cell less than cell log presuperscript 2 space 16 end cell row cell x squared plus x plus 4 end cell less than 16 row cell x squared plus x plus 4 minus 16 end cell less than 0 row cell x squared plus x minus 12 end cell less than 0 row cell open parentheses x minus 3 close parentheses open parentheses x plus 4 close parentheses end cell less than 0 row cell x minus 3 end cell less than 0 row x less than 3 row cell x plus 4 end cell less than 0 row x less than cell negative 4 end cell end table 

Menentukan daerah penyelesaian

Untuk x greater than negative 4 dan x less than 3, jika x equals 0 maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell x squared plus x minus 12 end cell less than 0 row cell 0 squared plus 0 minus 12 end cell less than 0 row cell negative 12 end cell less than cell 0 space open parentheses negatif close parentheses end cell end table 

Untuk x greater than 3, jika x equals 4 maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell x squared plus x minus 12 end cell greater than 0 row cell 4 squared plus 4 minus 12 end cell greater than 0 row cell 16 plus 4 minus 12 end cell greater than cell 0 space end cell row cell 20 minus 12 end cell greater than 0 row 8 greater than cell 0 space open parentheses positif close parentheses end cell end table 

Untuk x less than negative 4, jika x equals negative 5 maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell x squared plus x minus 12 end cell greater than 0 row cell open parentheses negative 5 close parentheses squared plus open parentheses negative 5 close parentheses minus 12 end cell greater than 0 row cell 25 minus 5 minus 12 end cell greater than 0 row cell 25 minus 17 end cell greater than 0 row 8 greater than cell space open parentheses positif close parentheses end cell end table 

Karena tanda pertidaksamaannya adalah kurang dari " less than " maka daerah penyelesaiannya adalah yang bernilai negatif.

Jadi, nilai x yang memenuhi adalah negative 4 less than x less than 3.

Oleh karena itu, jawaban yang benar adalah C.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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