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Tentukan himpunan penyelesaian dari tiap pertidaksamaan nilai mutlak berikut: f.

Pertanyaan

Tentukan himpunan penyelesaian dari tiap pertidaksamaan nilai mutlak berikut:

f. begin mathsize 14px style open vertical bar 3 x plus 2 close vertical bar greater or equal than open vertical bar x plus 6 close vertical bar end style 

Pembahasan Soal:

Penyelesaian dari pertidaksamaan tersebut adalah sebagai berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell vertical line 3 x plus 2 vertical line end cell greater or equal than cell vertical line x plus 6 vertical line end cell row cell open parentheses 3 x plus 2 close parentheses squared end cell greater or equal than cell open parentheses x plus 6 close parentheses squared end cell row cell left parenthesis 3 x plus 2 right parenthesis squared minus left parenthesis x plus 6 right parenthesis squared end cell greater or equal than cell 0 open square brackets open parentheses 3 x plus 2 close parentheses plus open parentheses x plus 6 close parentheses close square brackets open square brackets left parenthesis 3 x plus 2 right parenthesis minus left parenthesis x plus 6 right parenthesis close square brackets greater or equal than 0 end cell row cell open parentheses 4 x plus 8 close parentheses open parentheses 2 x minus 4 close parentheses end cell greater or equal than 0 row cell 4 left parenthesis x plus 2 right parenthesis 2 left parenthesis x minus 2 right parenthesis end cell greater or equal than 0 row cell left parenthesis x plus 2 right parenthesis left parenthesis x minus 2 right parenthesis end cell greater or equal than 0 row blank blank blank end table end style 

Jadi, himpunan penyelesaiannya adalah  begin mathsize 14px style open curly brackets x vertical line x less or equal than negative 2 space atau space x greater or equal than 2 comma x element of Real close curly brackets end style.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

E. Sri

Mahasiswa/Alumni Universitas Muhammadiyah Purworejo

Terakhir diupdate 30 Maret 2021

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Pertanyaan yang serupa

Himpunan penyelesaian dari pertidaksamaan  adalah ….

Pembahasan Soal:

Berdasarkan sifat undefined 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 2 x minus 1 close vertical bar end cell greater than cell vertical line 4 x plus 3 vertical line end cell row cell square root of open parentheses 2 x minus 1 close parentheses squared end root end cell greater than cell square root of open parentheses 4 x plus 3 close parentheses squared end root end cell row blank blank blank end table end style 

Kuadratkan kedua ruas

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses square root of open parentheses 2 x minus 1 close parentheses squared end root close parentheses squared end cell greater than cell open parentheses square root of open parentheses 4 x plus 3 close parentheses squared end root close parentheses squared end cell row cell 4 x squared minus 4 x plus 1 end cell greater than cell 16 x squared plus 24 x plus 9 end cell row cell 4 x squared minus 16 x squared minus 4 x minus 24 x plus 1 minus 9 end cell greater than 0 row cell negative 12 x squared minus 28 x minus 8 end cell greater than 0 row cell 12 x squared plus 28 x plus 8 end cell less than 0 row blank blank blank row blank blank blank row blank blank blank row blank blank blank end table end style 

Lakukan pemfaktoran

begin mathsize 14px style open parentheses 6 x plus 2 close parentheses open parentheses 2 x plus 4 close parentheses less than 0 end style 

Menentukan pembuat nol

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 6 x plus 2 end cell equals 0 row cell 6 x end cell equals cell negative 2 end cell row x equals cell negative 1 third end cell row blank blank blank row blank blank blank end table end style 

atau

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x plus 4 end cell equals 0 row cell 2 x end cell equals cell negative 4 end cell row x equals cell negative 2 end cell end table end style 

Maka diperoleh interval sebagai berikut

begin mathsize 14px style HP equals left curly bracket x vertical line minus 2 less than x less than negative 1 third right curly bracket end style 

Jadi, himpunan penyelesaian dari pertidaksamaan begin mathsize 14px style begin bold style vertical line 2 x minus 1 vertical line end style bold greater than bold vertical line bold 4 bold italic x bold plus bold 3 bold vertical line end style adalah A. begin mathsize 14px style begin bold style left curly bracket negative 2 less than x less than negative 1 third right curly bracket end style end style

0

Roboguru

Selesaikan setiap PtLSVNM berikut. d.

Pembahasan Soal:

Gunakan konsep penyelesaian pertidaksamaan bentuk open vertical bar f open parentheses x close parentheses close vertical bar greater than open vertical bar g open parentheses x close parentheses close vertical bar yaitu open square brackets f open parentheses x close parentheses plus g open parentheses x close parentheses close square brackets open square brackets f open parentheses x close parentheses minus g open parentheses x close parentheses close square brackets greater than 0.

Perhatikan perhitungan berikut.

table attributes columnalign right center left columnspacing 2px end attributes row cell 4 open vertical bar x minus 2 close vertical bar end cell greater than cell open vertical bar x close vertical bar end cell row cell open square brackets 4 open parentheses x minus 2 close parentheses plus x close square brackets open square brackets 4 open parentheses x minus 2 close parentheses minus x close square brackets end cell greater than 0 row cell open parentheses 4 x minus 8 plus x close parentheses open parentheses 4 x minus 8 minus x close parentheses end cell greater than 0 row cell open parentheses 5 x minus 8 close parentheses open parentheses 3 x minus 8 close parentheses end cell greater than 0 row cell x less than 8 over 5 end cell atau cell x greater than 8 over 3 end cell end table

Jadi, diperoleh penyelesaiannya adalah x less than 8 over 5 space atau space x greater than 8 over 3.

0

Roboguru

Himpunan penyelesaian dari pertidaksamaan  adalah ...

Pembahasan Soal:

Ingat sifat pertidaksamaan nilai mutlak

open vertical bar f open parentheses x close parentheses close vertical bar greater than open vertical bar g open parentheses x close parentheses close vertical bar left right double arrow open square brackets f open parentheses x close parentheses close square brackets squared greater than open square brackets g open parentheses x close parentheses close square brackets squared dan sifat 

a squared minus b squared equals open parentheses a plus b close parentheses open parentheses a minus b close parentheses

Sehingga penyelesaian dari open vertical bar x plus 3 close vertical bar greater than 2 open vertical bar x minus 3 close vertical bar dapat ditentukan dengan cara berikut.

space space space space open vertical bar x plus 3 close vertical bar greater than 2 open vertical bar x minus 3 close vertical bar left right double arrow open vertical bar x plus 3 close vertical bar greater than open vertical bar 2 x minus 6 close vertical bar left right double arrow open square brackets x plus 3 close square brackets squared greater than open square brackets 2 x minus 6 close square brackets squared left right double arrow open square brackets x plus 3 close square brackets squared minus open square brackets 2 x minus 6 close square brackets squared greater than 0 left right double arrow open parentheses x plus 3 cross times 2 x minus 6 close parentheses open parentheses x plus 3 minus 2 x plus 6 close parentheses greater than 0 left right double arrow open parentheses 3 x minus 3 close parentheses open parentheses negative x plus 9 close parentheses greater than 0 left right double arrow x equals 1 space atau space x equals 9

HP equals open curly brackets 1 less than x less than 9 close curly brackets

Dengan demikian jawaban yang tepat adalah D

 

0

Roboguru

Penyelesaian pertidaksamaan  adalah ....

Pembahasan Soal:

Gunakan konsep berikut.

open vertical bar f open parentheses x close parentheses close vertical bar greater or equal than g open parentheses x close parentheses left right double arrow f open parentheses x close parentheses less or equal than negative g open parentheses x close parentheses space atau space f open parentheses x close parentheses greater or equal 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 1 minus 2 x close vertical bar greater or equal than 3 minus x.

Perhatikan perhitungan berikut.

table row cell open vertical bar 1 minus 2 x close vertical bar greater or equal than 3 minus x end cell row cell table attributes columnalign right center left columnspacing 2px end attributes row cell 1 minus 2 x less or equal than negative open parentheses 3 minus x close parentheses end cell atau cell 1 minus 2 x greater or equal than 3 minus x end cell row cell 1 minus 2 x less or equal than negative 3 plus x end cell blank cell negative 2 x plus x greater or equal than 3 minus 1 end cell row cell negative 2 x minus x less or equal than negative 3 minus 1 end cell blank cell negative x greater or equal than 2 end cell row cell negative 3 x less or equal than negative 4 end cell blank cell negative x open parentheses negative 1 close parentheses less or equal than 2 open parentheses negative 1 close parentheses end cell row cell x greater or equal than fraction numerator negative 4 over denominator negative 3 end fraction end cell blank cell x less or equal than negative 2 end cell row cell x greater or equal than 4 over 3 end cell blank blank row cell x greater or equal than 1 1 third end cell blank blank end table end cell end table

Diperoleh penyelesaiannya adalah x less or equal than negative 2 atau x greater or equal than 1 1 third.

Jadi, jawaban yang tepat adalah C.

2

Roboguru

Himpunan penyelesaian dari pertidaksamaan  adalah

Pembahasan Soal:

  • Menyelesaikan pertidaksamaan nilai mutlak:

open vertical bar f left parenthesis x right parenthesis close vertical bar greater or equal than open vertical bar g left parenthesis x right parenthesis close vertical bar left right double arrow open square brackets f left parenthesis x right parenthesis close square brackets squared greater or equal than open square brackets g left parenthesis x right parenthesis close square brackets squared

sedemikian hingga:

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 2 x plus 1 close vertical bar end cell greater or equal than cell open vertical bar x minus 2 close vertical bar end cell row cell open square brackets 2 x plus 1 close square brackets squared end cell greater or equal than cell open square brackets x minus 2 close square brackets squared end cell row cell 4 x squared plus 4 x plus 1 end cell greater or equal than cell x squared minus 4 x plus 4 end cell row cell 3 x squared plus 8 x minus 3 end cell greater or equal than 0 end table

Pembuat nol:

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 x squared plus 8 x minus 3 end cell equals 0 row cell open parentheses x plus 3 close parentheses open parentheses 3 x minus 1 close parentheses end cell equals 0 end table

x equals negative 3 space atau space x equals 1 third

Garis bilangan:

Jadi, himpunan penyelesaian dari pertidaksamaan tersebut adalah open curly brackets x vertical line x less or equal than negative 3 space atau space x greater or equal than 1 third comma space x element of R close curly brackets.

Oleh karena itu, jawaban yang benar adalah A.

0

Roboguru

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