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Pertanyaan

Selesaikan open vertical bar x minus 7 close vertical bar equals negative open vertical bar 2 x plus 1 close vertical bar.

Pembahasan Soal:

Ingat kembali sifat dari nilai mutlak.

open vertical bar A close vertical bar equals open vertical bar negative A close vertical bar equals A

Persamaan yang diketahui adalah

open vertical bar x minus 7 close vertical bar equals negative open vertical bar 2 x plus 1 close vertical bar

Dari persamaan di atas, hasil dari open vertical bar x minus 7 close vertical bar adalah negatif open vertical bar 2 x plus 1 close vertical bar. Hal ini tidak mungkin, karena nilai yang dimutlakkan selalu positif.

Dengan demikian, penyelesaian dari open vertical bar x minus 7 close vertical bar equals negative open vertical bar 2 x plus 1 close vertical bar itu tidak ada .

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Rahma

Mahasiswa/Alumni Institut Teknologi Bandung

Terakhir diupdate 11 Agustus 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Carilah nilai  yang memenuhi persamaan nilai mutlak berikut ini. b.

Pembahasan Soal:

Ingat bahwa:

Nilai mutlak memiliki sifat open vertical bar a close vertical bar squared equals open parentheses a close parentheses squared.

Sehingga diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar fraction numerator 2 x minus 3 over denominator 3 end fraction close vertical bar end cell equals cell open vertical bar 1 third x minus 1 close vertical bar end cell row cell open parentheses fraction numerator 2 x minus 3 over denominator 3 end fraction close parentheses squared end cell equals cell open parentheses 1 third x minus 1 close parentheses squared end cell row cell fraction numerator 4 x squared minus 12 x plus 9 over denominator 9 end fraction end cell equals cell 1 over 9 x squared minus 2 over 3 x plus 1 end cell row cell 4 x squared minus 12 x plus 9 end cell equals cell 9 open parentheses 1 over 9 x squared minus 2 over 3 x plus 1 close parentheses end cell row cell 4 x squared minus 12 x plus 9 end cell equals cell x squared minus 6 x plus 9 end cell row cell 4 x squared minus x squared minus 12 x plus 6 x plus 9 minus 9 end cell equals 0 row cell 3 x squared minus 6 x end cell equals 0 row cell 3 x open parentheses x minus 2 close parentheses end cell equals 0 row cell 3 x end cell equals 0 row x equals 0 row blank blank atau row cell x minus 2 end cell equals 0 row x equals 2 end table 

Dengan demikian, nilai x yang memenuhi persamaan nilai mutlak open vertical bar fraction numerator 2 x minus 3 over denominator 3 end fraction close vertical bar equals open vertical bar 1 third x minus 1 close vertical bar adalah x equals 0 space atau space x equals 2.

0

Roboguru

Tentukan nilai  yang memenuhi persamaan nilai mutlak sebagai berikut:

Pembahasan Soal:

Dengan menguadratkan kedua ruas, perhatikan perhitungan berikut:

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 2 x minus 5 close vertical bar end cell equals cell open vertical bar 7 minus 2 x close vertical bar end cell row cell left parenthesis 2 x minus 5 right parenthesis squared end cell equals cell left parenthesis 7 minus 2 x right parenthesis squared space end cell row cell left parenthesis 2 x minus 5 right parenthesis squared minus left parenthesis 7 minus 2 x right parenthesis squared end cell equals 0 row cell left parenthesis left parenthesis 2 x minus 5 right parenthesis plus left parenthesis 7 minus 2 x right parenthesis right parenthesis space left parenthesis left parenthesis 2 x minus 5 right parenthesis minus left parenthesis 7 minus 2 x right parenthesis right parenthesis end cell equals 0 row cell 2 space left parenthesis 4 x minus 12 right parenthesis end cell equals 0 end table 

Maka diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell 4 x minus 12 end cell equals 0 row cell 4 x end cell equals 12 row x equals cell 12 over 4 end cell row x equals 3 end table 

Dengan demikian, nilai x yang memenuhi persamaan nilai mutlak open vertical bar 2 x minus 5 close vertical bar equals open vertical bar 7 minus 2 x close vertical bar adalah x equals 3

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Roboguru

Tentukanlah HP

Pembahasan Soal:

Ingatlah aturan nilai mutlak!

open vertical bar x close vertical bar equals open curly brackets table attributes columnalign left end attributes row cell negative x rightwards arrow x less or equal than 0 end cell row cell x rightwards arrow x greater than 0 end cell end table close

Maka,

a. open vertical bar x plus 1 close vertical bar

open vertical bar x plus 1 close vertical bar equals x plus 1

table attributes columnalign right center left columnspacing 0px end attributes row cell x plus 1 end cell greater or equal than cell thin space 0 end cell row cell x plus 1 minus 1 end cell greater or equal than cell thin space 0 minus 1 end cell row x greater or equal than cell thin space minus 1 end cell end table

atau

open vertical bar x plus 1 close vertical bar equals negative open parentheses x plus 1 close parentheses

table attributes columnalign right center left columnspacing 0px end attributes row cell x plus 1 end cell less than 0 row cell x plus 1 minus 1 end cell less than cell 0 minus 1 end cell row x less than cell negative 1 end cell end table

b.open vertical bar x minus 4 close vertical bar

open vertical bar x minus 4 close vertical bar equals x minus 4

table attributes columnalign right center left columnspacing 0px end attributes row cell x minus 4 end cell greater or equal than cell thin space 0 end cell row cell x minus 4 plus 4 end cell greater or equal than cell thin space 0 plus 4 end cell row x greater or equal than cell thin space 4 end cell end table

atau

table attributes columnalign right center left columnspacing 0px end attributes row cell x minus 4 end cell less than 0 row cell x minus 4 plus 4 end cell less than cell 0 plus 4 end cell row x less than 4 end table

Di dapatkan interval x less than negative 1negative 1 less or equal than thin space x less than 4, dan x greater or equal than thin space 4.

Untuk interval x less than negative 1

table attributes columnalign right center left columnspacing 0px end attributes row cell negative open parentheses x plus 1 close parentheses end cell equals cell negative open parentheses x minus 4 close parentheses end cell row cell negative x minus 1 end cell equals cell negative x plus 4 end cell row cell negative x minus 1 plus 1 end cell equals cell negative x plus 4 plus 1 end cell row cell negative x end cell equals cell negative x plus 5 end cell row cell negative x plus x end cell equals cell negative x plus 5 plus x end cell row 0 equals cell 5 space left parenthesis tidak space memenuhi right parenthesis end cell end table

Untuk interval negative 1 less or equal than thin space x less than 4,

table attributes columnalign right center left columnspacing 0px end attributes row cell x plus 1 end cell equals cell negative open parentheses x minus 4 close parentheses end cell row cell x plus 1 end cell equals cell negative x plus 4 end cell row cell x plus 1 minus 1 end cell equals cell negative x plus 4 minus 1 end cell row x equals cell negative x plus 3 end cell row cell x plus x end cell equals cell negative x plus 3 plus x end cell row cell 2 x end cell equals 3 row cell fraction numerator 2 x over denominator 2 end fraction end cell equals cell 3 over 2 end cell row x equals cell 3 over 2 end cell row blank blank blank end table

Di dapatkan negative 1 less or equal than thin space x less than 4 dan x equals 3 over 2.

Untuk interval x greater or equal than thin space 4

table attributes columnalign right center left columnspacing 0px end attributes row cell x plus 1 end cell equals cell x minus 4 end cell row cell x plus 1 minus x end cell equals cell x minus 4 minus x end cell row 1 equals cell negative 4 open parentheses tidak space memenuhi close parentheses end cell end table



Jadi, himpunan penyelesaian dari open vertical bar x plus 1 close vertical bar equals open vertical bar x minus 4 close vertical bar adalah x equals 3 over 2.

0

Roboguru

Nilai  yang memenuhi persamaan  adalah ...

Pembahasan Soal:

Nilai mutlak atau bisa juga disebut dengan modulus merupakan nilai suatu bilangan riil tanpa adanya tanda tambah (+) atau kurang (–).

Nilai mutlak mempunyai sifat: open vertical bar a close vertical bar squared equals a squared, sehingga:

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 5 x minus 1 close vertical bar minus open vertical bar x plus 8 close vertical bar end cell equals 0 row cell open vertical bar 5 x minus 1 close vertical bar end cell equals cell open vertical bar x plus 8 close vertical bar end cell row cell left parenthesis open vertical bar 5 x minus 1 close vertical bar right parenthesis squared end cell equals cell left parenthesis open vertical bar x plus 8 close vertical bar right parenthesis squared space end cell row cell open parentheses 5 x minus 1 close parentheses squared end cell equals cell open parentheses x plus 8 close parentheses squared end cell row cell 25 x squared minus 10 x plus 1 end cell equals cell x squared plus 16 x plus 64 end cell row cell 24 x squared minus 26 x minus 63 end cell equals 0 row cell left parenthesis 4 x minus 9 right parenthesis left parenthesis 6 x plus 7 right parenthesis end cell equals 0 row cell 4 x minus 9 end cell equals 0 row cell 4 x end cell equals 9 row x equals cell 9 over 4 end cell row blank blank atau row cell 6 x plus 7 end cell equals 0 row cell 6 x end cell equals cell negative 7 end cell row x equals cell negative 7 over 6 end cell end table 

Oleh karena itu, jawaban yang tepat adalah A.

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Roboguru

Hasil jumlah semua nilai  yang merupakan penyelesaian dari persamaan  adalah ...

Pembahasan Soal:

Bentuk ini 3 open vertical bar n plus 3 close vertical bar equals open vertical bar 8 minus 2 n close vertical bar dapat kita tinjau dalam dua arah sebagai berikut.

Tinjauan pertama:


3 left parenthesis n plus 3 right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell 8 minus 2 n end cell row cell negative open parentheses 8 minus 2 n close parentheses end cell end table close


table attributes columnalign right center left columnspacing 0px end attributes row cell 3 n plus 9 end cell equals cell 8 minus 2 n end cell row cell 3 n bold plus bold 2 bold italic n end cell equals cell 8 bold minus bold 9 end cell row cell 5 n end cell equals cell negative 1 end cell row n equals cell negative 1 fifth end cell end table


table attributes columnalign right center left columnspacing 0px end attributes row cell 3 n plus 9 end cell equals cell negative open parentheses 8 minus 2 n close parentheses end cell row cell 3 n plus 9 end cell equals cell 2 n minus 8 end cell row cell 3 n minus bold 2 bold italic n end cell equals cell negative 8 bold minus bold 9 end cell row n equals cell negative 17 end cell end table


Tinjauan kedua:


8 minus 2 n equals open curly brackets table attributes columnalign left end attributes row cell 3 left parenthesis n plus 3 right parenthesis end cell row cell negative 3 open parentheses n plus 3 close parentheses end cell end table close


table attributes columnalign right center left columnspacing 0px end attributes row cell 8 minus 2 n end cell equals cell 3 n plus 9 end cell row cell negative 2 n bold minus bold 3 bold italic n end cell equals cell 9 bold minus bold 8 end cell row cell negative 5 n end cell equals 1 row n equals cell negative 1 fifth end cell end table


table attributes columnalign right center left columnspacing 0px end attributes row cell 8 minus 2 n end cell equals cell negative open parentheses 3 n plus 9 close parentheses end cell row cell 8 minus 2 n end cell equals cell negative 3 n minus 9 end cell row cell negative 2 n plus 3 n end cell equals cell negative 9 minus 8 end cell row n equals cell negative 17 end cell end table


Jadi, HP equals open curly brackets negative 17 comma space minus 1 fifth close curly brackets, maka umlah semua nilai n adalah:


table attributes columnalign right center left columnspacing 0px end attributes row cell negative 17 plus open parentheses negative 1 fifth close parentheses end cell equals cell negative 17 minus 0 comma 2 end cell row blank equals cell negative 17 comma 2 end cell end table


Oleh karena itu, jawaban yang benar adalah A.

0

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