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Himpunan penyelesaian dari  adalah interval . Nilai  adalah ....

Pertanyaan

Himpunan penyelesaian dari begin mathsize 14px style open vertical bar x minus 1 close vertical bar less than 6 over x end style adalah interval begin mathsize 14px style left parenthesis straight a comma space straight b right parenthesis end style. Nilai begin mathsize 14px style 3 a plus 2 b end style adalah ....

  1. 0

  2. 2

  3. 4

  4. 6

  5. 12

Pembahasan Soal:

Dengan menerapkan konsep pertidaksamaan nilai mutlak:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar straight x minus 1 close vertical bar end cell less than cell 6 over straight x end cell row cell left parenthesis straight x minus 1 right parenthesis squared end cell less than cell left parenthesis 6 over straight x right parenthesis squared end cell row cell space left parenthesis straight x minus 1 right parenthesis squared minus left parenthesis 6 over straight x right parenthesis squared end cell less than 0 row cell left parenthesis straight x minus 1 minus 6 over straight x right parenthesis left parenthesis straight x minus 1 plus 6 over straight x right parenthesis end cell less than 0 row cell left parenthesis fraction numerator straight x squared minus straight x minus 6 over denominator straight x end fraction right parenthesis left parenthesis fraction numerator straight x squared minus straight x plus 6 over denominator straight x end fraction right parenthesis end cell less than 0 row cell left parenthesis fraction numerator left parenthesis straight x minus 3 right parenthesis left parenthesis straight x plus 2 right parenthesis over denominator straight x end fraction right parenthesis left parenthesis fraction numerator straight x squared minus straight x plus 6 over denominator straight x end fraction right parenthesis end cell less than 0 row blank blank blank row blank blank blank end table end style  

Dari pertidaksamaan di atas, maka nilai begin mathsize 14px style x end style pembuat Nol adalah begin mathsize 14px style straight x equals 3 space union straight x equals space minus 2 end style dengan syarat begin mathsize 14px style x not equal to 0 end style.

Dengan uji coba pada garis bilangan:

begin mathsize 14px style table row cell plus plus plus end cell cell negative negative negative space space space space space minus negative negative end cell blank cell plus plus plus end cell row cell space space space space space space space space space minus 2 end cell cell space space 0 end cell 3 blank end table end style 

Himpunan penyelesaian adalah himpunanan bernilai negatif yaitu begin mathsize 14px style 0 less than x less than 3 end style, untuk begin mathsize 14px style negative 2 less than x less than 0 end style (tidak memenuhi) karena akan membuat begin mathsize 14px style 6 over x end style bernilai negatif.

Kesimpulannya:

begin mathsize 14px style straight a equals 0 comma space straight b equals space 3 end style, sehingga diperoleh:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 3 straight a plus 2 straight b end cell equals cell 3 times 0 plus 2 times 3 end cell row blank equals cell 0 plus 6 end cell row blank equals 6 end table end style 

Jadi, Nilai dari begin mathsize 14px style 3 a plus 2 b end style adalah 6.

Oleh karena itu, Jawaban yang benar adalah D

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

R. RGFLLIMA

Terakhir diupdate 30 April 2021

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Pertanyaan yang serupa

Diketahui pertidaksamaan nilai mutlak , maka nilai  adalah ...

Pembahasan Soal:

Pembuat nol nilai mutlak tersebut adalah sebagai berikut.

open vertical bar 4 x plus 2 close vertical bar equals 0 space left right double arrow space 4 x plus 2 equals 0 space left right double arrow space x equals negative 2 over 4 equals negative 1 half

Untuk interval x less or equal than negative 1 half

open vertical bar 4 x plus 2 close vertical bar equals negative open parentheses 4 x plus 2 close parentheses

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 4 x plus 2 close vertical bar end cell less or equal than cell 2 x plus 6 end cell row cell negative open parentheses 4 x plus 2 close parentheses end cell less or equal than cell 2 x plus 6 end cell row cell negative 4 x minus 2 end cell less or equal than cell 2 x plus 6 end cell row cell negative 6 x end cell less or equal than 8 row x greater or equal than cell negative 8 over 6 end cell row x greater or equal than cell negative 4 over 3 end cell end table

Pertidaksamaan open vertical bar 4 x plus 2 close vertical bar less or equal than 2 x plus 6 untuk interval x less or equal than negative 1 half mempunyai penyelesaian negative 4 over 3 less or equal than x less or equal than negative 1 half 

Untuk interval x greater or equal than negative 1 half

open vertical bar 4 x plus 2 close vertical bar equals 4 x plus 2

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 4 x plus 2 close vertical bar end cell less or equal than cell 2 x plus 6 end cell row cell 4 x plus 2 end cell less or equal than cell 2 x plus 6 end cell row cell 4 x minus 2 x end cell less or equal than cell 6 minus 2 end cell row cell 2 x end cell less or equal than 4 row x less or equal than 2 end table

Pertidaksamaan open vertical bar 4 x plus 2 close vertical bar less or equal than 2 x plus 6 untuk interval x greater or equal than negative 1 half mempunyai penyelesaian negative 1 half less or equal than x less or equal than 2

Gabungan penyelesaian, yaitu

negative 4 over 3 less or equal than x less or equal than 2

Dengan demikian, penyelesaian pertidaksamaan nilai mutlak open vertical bar 4 x plus 2 close vertical bar less or equal than 2 x plus 6 adalah negative 4 over 3 less or equal than x less or equal than 2

0

Roboguru

Tentukan nilai  yang memenuhi pertidaksamaan berikut. 7.

Pembahasan Soal:

Perhatikan sifat pertidaksamaan nilai mutlak berikut.

open vertical bar x close vertical bar less than y rightwards double arrow x squared less than y squared 

Sehingga dapat ditentukan penyelesaian dari pertidaksamaan nilai mutlak di atas sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 6 x minus 8 close vertical bar end cell less than cell 2 x minus 1 end cell row cell open vertical bar 6 x minus 8 close vertical bar squared end cell less than cell open parentheses 2 x minus 1 close parentheses squared end cell row cell open parentheses 6 x minus 8 close parentheses squared end cell less than cell open parentheses 2 x minus 1 close parentheses squared end cell row cell 36 x squared minus 96 x plus 64 end cell less than cell 4 x squared minus 4 x plus 1 end cell row cell 32 x squared minus 92 x plus 63 end cell less than 0 row cell open parentheses 4 x minus 7 close parentheses open parentheses 8 x minus 9 close parentheses end cell less than 0 end table  

Berdasarkan bentuk terakhir yang diperoleh di atas, maka dapat ditentukan pembuat nol pertidaksamaan di atas sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell 4 x minus 7 end cell equals cell 0 space atau space 8 x minus 9 equals 0 end cell row cell 4 x end cell equals cell 7 space atau space 8 x equals 9 end cell row x equals cell 7 over 4 space atau space x equals 9 over 8 end cell row x equals cell 1 3 over 4 space atau space x equals 1 1 over 8 end cell end table 

Pembuat nol di atas membagi garis bilangan menjadi tiga daerah. Selanjutnya diuji masing-masing titik pada masing-masing daerah yang terbentuk untuk menentukan tanda positif atau negatif, seperti ditunjukkan berikut.

table attributes columnalign right center left columnspacing 0px end attributes row x equals cell 0 rightwards double arrow open parentheses 4 x minus 7 close parentheses open parentheses 8 x minus 9 close parentheses equals open parentheses 4 times 0 minus 7 close parentheses open parentheses 8 times 0 minus 9 close parentheses equals plus 63 end cell row x equals cell 10 over 8 rightwards double arrow open parentheses 4 x minus 7 close parentheses open parentheses 8 x minus 9 close parentheses equals open parentheses 4 times 10 over 8 minus 7 close parentheses open parentheses 8 times 10 over 8 minus 9 close parentheses equals negative 2 end cell row x equals cell 2 rightwards double arrow open parentheses 4 x minus 7 close parentheses open parentheses 8 x minus 9 close parentheses equals open parentheses 4 times 2 minus 7 close parentheses open parentheses 8 times 2 minus 9 close parentheses equals plus 7 end cell end table 

Karena tanda ketaksamaannya adalah kurang dari (less than), maka daerah yang bertanda negatif yang menjadi penyelesaiannya, seperti ditunjukkan pada gambar di atas.

Dengan demikian, nilai x yang memenuhi pertidaksamaan yang diberikan di atas dapat ditunjukkan dalam himpunan penyelesaian pertidaksamaan nilai mutlak yaitu open curly brackets right enclose x 9 over 8 less than x less than 7 over 4 close curly brackets.

0

Roboguru

Penyelesaian pertidaksamaan  adalah ....

Pembahasan Soal:

Gunakan konsep berikut.

open vertical bar f open parentheses x close parentheses close vertical bar less than g open parentheses x close parentheses left right double arrow negative g open parentheses x close parentheses less than f open parentheses x close parentheses less than g open parentheses x close parentheses dan g open parentheses x close parentheses greater or equal than 0

Menggunakan konsep di atas, akan dicari penyelesaian pertidaksamaan open vertical bar 2 x minus 1 close vertical bar less than x plus 1.

Perhatikan perhitungan berikut.

table row cell open vertical bar 2 x minus 1 close vertical bar less than x plus 1 end cell row cell negative open parentheses x plus 1 close parentheses less than 2 x minus 1 less than x plus 1 end cell row cell negative x minus 1 less than 2 x minus 1 less than x plus 1 end cell row cell table attributes columnalign right center left columnspacing 2px end attributes row cell negative x minus 1 less than 2 x minus 1 end cell cell space dan space end cell cell 2 x minus 1 less than x plus 1 end cell row cell negative x minus 2 x less than negative 1 plus 1 end cell dan cell 2 x minus x less than 1 plus 1 end cell row cell negative 3 x less than 0 end cell dan cell x less than 2 end cell row cell x greater than fraction numerator 0 over denominator negative 3 end fraction end cell blank blank row cell x greater than 0 end cell blank blank end table end cell end table

Diperoleh nilai x yang memenuhi adalah x greater than 0 dan x less than 2, jika digabungkan akan menjadi 0 less than x less than 2.

Jadi, jawaban yang tepat adalah C.

0

Roboguru

Himpunan penyelesaian dari adalah…

Pembahasan Soal:

Ingat sifat mutlak: open vertical bar f left parenthesis x right parenthesis close vertical bar less than a rightwards arrow negative a less than f left parenthesis x right parenthesis less than a

Maka,

vertical line 2 x minus 5 vertical line less than 3 rightwards arrow negative 3 less than 2 x minus 5 less than 3 space space space space space space space space space space space space space space space space space space space space space space space minus 3 plus 5 less than 2 x less than 3 plus 5 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space 2 less than 2 x less than 8 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space 1 less than x less than 4

Jadi, himpunan penyelesaian pertidaksamaan tersebut adalah 1 less than x less than 4.

0

Roboguru

Nilai  yang memenuhi  adalah ...

Pembahasan Soal:

Ingat bahwa:

Nilai mutlak memiliki sifat jika open vertical bar f left parenthesis x right parenthesis close vertical bar less than a maka negative a less than f left parenthesis x right parenthesis less than a. Selain itu bentuk open parentheses a plus b close parentheses squared equals a squared plus 2 a b plus b squared dan bentuk akar memiliki sifat open parentheses square root of a close parentheses squared equals a.

Sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar x plus 3 close vertical bar end cell less than cell square root of 9 minus x squared end root end cell row cell negative open parentheses square root of 9 minus x squared end root close parentheses end cell less than cell x plus 3 less than square root of 9 minus x squared end root end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses negative square root of 9 minus x squared end root close parentheses squared end cell less than cell open parentheses x plus 3 close parentheses squared end cell row cell 9 minus x squared end cell less than cell x squared plus 6 x plus 9 end cell row cell negative x squared minus x squared minus 6 x plus 9 minus 9 end cell less than 0 row cell negative 2 x squared minus 6 x end cell less than 0 row cell 2 x squared plus 6 x end cell greater than 0 row cell 2 x open parentheses x plus 3 close parentheses end cell greater than 0 row cell 2 x end cell greater than 0 row x greater than 0 row blank blank atau row cell straight x plus 3 end cell greater than 0 row straight x greater than cell negative 3 end cell end table

   table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses x plus 3 close parentheses squared end cell less than cell open parentheses square root of 9 minus x squared end root close parentheses squared end cell row cell x squared plus 6 x plus 9 end cell less than cell 9 minus x squared end cell row cell x squared plus x squared plus 6 x plus 9 minus 9 end cell less than 0 row cell 2 x squared plus 6 x end cell less than 0 row cell 2 x left parenthesis x plus 3 right parenthesis end cell less than 0 row cell 2 x end cell less than 0 row x less than 0 row blank blank atau row cell x plus 3 end cell less than 0 row x less than cell negative 3 end cell end table

Perhatikan syarat bentuk akar dimana square root of a comma space a greater or equal than 0, sehingga bentuk akar pada soal memiliki syarat

table attributes columnalign right center left columnspacing 0px end attributes row cell 9 minus x squared end cell greater or equal than 0 row cell negative x squared end cell greater or equal than cell negative 9 end cell row cell x squared end cell greater or equal than 9 row cell negative 3 end cell less or equal than cell x less or equal than 3 end cell end table

Lakukan uji titik untuk menentukan daerah penyelesaiannya.

Misal x equals negative 2

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar negative 2 plus 3 close vertical bar end cell less than cell square root of 9 minus left parenthesis negative 2 right parenthesis squared end root end cell row cell open vertical bar 1 close vertical bar end cell less than cell square root of 5 end cell row 1 less than cell square root of 5 horizontal ellipsis open parentheses benar close parentheses end cell end table

Misal x equals 1

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 1 plus 3 close vertical bar end cell less than cell square root of 9 minus left parenthesis 1 right parenthesis squared end root end cell row cell open vertical bar 4 close vertical bar end cell less than cell square root of 8 end cell row 4 less than cell square root of 8 horizontal ellipsis open parentheses salah close parentheses end cell end table

Berdasarkan pemisalan dan syarat daerah penyelesaian pertidaksamaan di atas diperoleh bahwa pertidaksamaan open vertical bar x plus 3 close vertical bar less than square root of 9 minus x squared end root adalah negative 3 less than x less than 0.

Jadi, jawaban yang tepat adalah E.

0

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