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Tentukan nilai  yang memenuhi persamaan berikut :

Pertanyaan

Tentukan nilai x yang memenuhi persamaan berikut :

open vertical bar x minus 3 close vertical bar minus open vertical bar 2 x minus 8 close vertical bar equals 5

Pembahasan Soal:

Definisi dari nilai mutlak adalah :

open vertical bar x close vertical bar equals open curly brackets table row cell negative x comma space x less than 0 end cell row cell x comma space x greater or equal than 0 end cell end table close

Dari soal diperoleh persamaan berikut :

open vertical bar x minus 3 close vertical bar minus open vertical bar 2 x minus 8 close vertical bar equals 5

Berdasarkan definisi nilai mutlak :

open vertical bar x minus 3 close vertical bar equals open curly brackets table row cell negative open parentheses x minus 3 close parentheses comma space x minus 3 less than 0 end cell row cell x minus 3 comma space x minus 3 greater or equal than 0 end cell end table close

open vertical bar x minus 3 close vertical bar equals open curly brackets table row cell 3 minus x comma space x less than 3 end cell row cell x minus 3 comma space x greater or equal than 3 end cell end table close

dan

open vertical bar 2 x minus 8 close vertical bar equals open curly brackets table row cell negative open parentheses 2 x minus 8 close parentheses comma space 2 x minus 8 less than 0 end cell row cell 2 x minus 8 comma space 2 x minus 8 greater or equal than 0 end cell end table close

open vertical bar 2 x minus 8 close vertical bar equals open curly brackets table row cell 8 minus 2 x comma space x less than 4 end cell row cell 2 x minus 8 comma space x greater or equal than 4 end cell end table close

 

Berdasarkan batas-batas nilai x diatas, kita gunakan garis bilangan di bawah ini :

 

Berdasarkan garis bilangan diatas, kita bagi menjadi 3 daerah yaitu :

Jika x less than 3, maka :

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar x minus 3 close vertical bar minus open vertical bar 2 x minus 8 close vertical bar end cell equals 5 row cell negative left parenthesis x minus 3 right parenthesis plus 2 x minus 8 end cell equals 5 row cell negative x plus 3 plus 2 x minus 8 end cell equals 5 row x equals 10 end table

Karena syarat yang harus terpenuhi adalah x less than 3 maka nilai x equals 10 tidak memenuhi.

Jika 3 less or equal than x less or equal than 4, maka :

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar x minus 3 close vertical bar minus open vertical bar 2 x minus 8 close vertical bar end cell equals 5 row cell x minus 3 plus 2 x minus 8 end cell equals 5 row cell 3 x end cell equals 16 row x equals cell 16 over 3 end cell end table

Karena syarat yang harus terpenuhi adalah 3 less or equal than x less or equal than 4 maka nilai x equals 16 over 3 tidak memenuhi.

Jika x greater than 3, maka :

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar x minus 3 close vertical bar minus open vertical bar 2 x minus 8 close vertical bar end cell equals 5 row cell x minus 3 minus left parenthesis 2 x minus 8 right parenthesis end cell equals 5 row cell x minus 3 minus 2 x plus 8 end cell equals 5 row x equals 0 end table

Karena syarat yang harus terpenuhi adalah x greater than 3 maka nilai x equals 0 tidak memenuhi.

Dengan demikian, tidak ada nilai x yang memenuhi.

 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

M. Gandhi

Mahasiswa/Alumni Universitas Negeri Yogyakarta

Terakhir diupdate 12 Agustus 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Hitunglah nilai  yang memenuhi persamaan berikut.

Pembahasan Soal:

Ingat definisi nilai mutlak:

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

Maka dapat kita peroleh:

open vertical bar 2 x minus 1 close vertical bar equals open curly brackets table attributes columnalign left end attributes row cell 2 x minus 1 space space space space space space space space comma space x greater or equal than 1 half end cell row cell negative left parenthesis 2 x minus 1 right parenthesis space space comma space x less than 1 half end cell end table close 

dan

open vertical bar 3 x minus 2 close vertical bar equals open curly brackets table attributes columnalign left end attributes row cell 3 x minus 2 space space space space space space space space comma space x greater or equal than 2 over 3 end cell row cell negative left parenthesis 3 x minus 2 right parenthesis space space comma space x less than 2 over 3 end cell end table close 

Berdasarkan definisi nilai mutlak tersebut, maka terdapat tiga kemungkinan kasus penyelesaian yaitu:

1. Untuk x less than 1 half:

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 2 x minus 1 close vertical bar plus open vertical bar 3 x minus 2 close vertical bar end cell equals 5 row cell negative left parenthesis 2 x minus 1 right parenthesis plus left parenthesis negative left parenthesis 3 x minus 2 right parenthesis right parenthesis end cell equals 5 row cell negative 2 x plus 1 minus 3 x plus 2 end cell equals 5 row cell negative 5 x plus 3 end cell equals 5 row cell negative 5 x end cell equals 2 row x equals cell negative 2 over 5 end cell end table 

Diperoleh x equals negative 2 over 5, artinya bahwa x less than 1 half (memenuhi). Sehingga x equals negative 2 over 5 merupakan solusi.

2. Untuk 1 half less or equal than x less than 2 over 3:

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 2 x minus 1 close vertical bar plus open vertical bar 3 x minus 2 close vertical bar end cell equals 5 row cell 2 x minus 1 plus left parenthesis negative left parenthesis 3 x minus 2 right parenthesis right parenthesis end cell equals 5 row cell 2 x minus 1 minus 3 x plus 2 end cell equals 5 row cell negative x plus 1 end cell equals 5 row cell negative x end cell equals 4 row x equals cell negative 4 end cell end table 

Nilai x equals negative 4 tidak memenuhi batasan 1 half less or equal than x less than 2 over 3, maka x equals negative 4 bukan merupakan solusi.

3. Untuk x greater or equal than 2 over 3:

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 2 x minus 1 close vertical bar plus open vertical bar 3 x minus 2 close vertical bar end cell equals 5 row cell 2 x minus 1 plus 3 x minus 2 end cell equals 5 row cell 5 x minus 3 end cell equals 5 row cell 5 x end cell equals 2 row x equals cell 2 over 5 end cell end table 

Nilai x equals 2 over 5 tidak memenuhi batasan x greater or equal than 2 over 3, maka x equals 2 over 5 bukan merupakan solusi.

Dengan demikian, nilai x yang memenuhi persamaan open vertical bar 2 x minus 1 close vertical bar plus open vertical bar 3 x minus 2 close vertical bar equals 5 adalah x equals negative 2 over 5

0

Roboguru

Tentukan himpunan penyelesaiannya!

Pembahasan Soal:

Ingat kembali definisi nilai mutlak untuk setiap bilangan real x adalah sebagai berikut.

open vertical bar f open parentheses x close parentheses close vertical bar equals open curly brackets table attributes columnalign center left end attributes row cell plus f open parentheses x close parentheses comma end cell cell untuk space f open parentheses x close parentheses greater or equal than 0 end cell row cell negative f open parentheses x close parentheses comma end cell cell untuk space f open parentheses x close parentheses less than 0 end cell end table close 

Akan ditentukan himpunan penyelesaian dari open vertical bar 3 x plus 1 close vertical bar plus open vertical bar 2 x minus 3 close vertical bar equals 5.

Terlebih dahulu jabarkan open vertical bar 3 x plus 1 close vertical bar plus open vertical bar 2 x minus 3 close vertical bar menggunakan definisi nilai mutlak diperoleh.

begin mathsize 12px style table attributes columnalign right center left columnspacing 2px end attributes row blank blank cell open vertical bar 3 x plus 1 close vertical bar plus open vertical bar 2 x minus 3 close vertical bar end cell row blank equals cell open curly brackets table attributes columnalign center left end attributes row cell open parentheses 3 x plus 1 close parentheses plus open parentheses 2 x minus 3 close parentheses comma space end cell cell untuk space 3 x plus 1 greater or equal than 0 space dan space 2 x minus 3 greater or equal than 0 end cell row cell open parentheses 3 x plus 1 close parentheses minus open parentheses 2 x minus 3 close parentheses comma end cell cell untuk space 3 x plus 1 greater or equal than 0 space dan space 2 x minus 3 less than 0 end cell row cell negative open parentheses 3 x plus 1 close parentheses plus open parentheses 2 x minus 3 close parentheses comma end cell cell untuk space 3 x plus 1 less than 0 space dan space 2 x minus 3 greater or equal than 0 end cell row cell negative open parentheses 3 x plus 1 close parentheses minus open parentheses 2 x minus 3 close parentheses comma end cell cell untuk space 3 x plus 1 less than 0 space dan space 2 x minus 3 less than 0 end cell end table close end cell row blank equals cell open curly brackets table attributes columnalign center left end attributes row cell 3 x plus 1 plus 2 x minus 3 comma space end cell cell untuk space x greater or equal than negative 1 third space dan space x greater or equal than 3 over 2 end cell row cell 3 x plus 1 minus 2 x plus 3 comma end cell cell untuk space x greater or equal than negative 1 third space dan space x less than 3 over 2 end cell row cell negative 3 x minus 1 plus 2 x minus 3 comma end cell cell untuk space x less than negative 1 third space dan space x greater or equal than 3 over 2 end cell row cell negative 3 x minus 1 minus 2 x plus 3 comma end cell cell untuk space x less than negative 1 third space dan space x less than 3 over 2 end cell end table close end cell row blank equals cell open curly brackets table attributes columnalign center left end attributes row cell 5 x minus 2 comma space end cell cell untuk space x greater or equal than 3 over 2 end cell row cell x plus 4 comma end cell cell untuk space minus 1 third less or equal than x less than 3 over 2 end cell row cell negative x minus 4 comma end cell cell untuk space x less than negative 1 third space dan space x greater or equal than 3 over 2 space left parenthesis tidak space memenuhi right parenthesis end cell row cell negative 5 x plus 2 comma end cell cell untuk space x less than negative 1 third end cell end table close end cell end table end style

Diperoleh yang memenuhi syarat adalah 5 x minus 2 untuk x greater or equal than 3 over 2x plus 4 untuk negative 1 third less or equal than x less than 3 over 2, dan negative 5 x plus 2 untuk x less than negative 1 third.

Nilai x pada 5 x minus 2 untuk x greater or equal than 3 over 2.

table attributes columnalign right center left columnspacing 2px end attributes row cell open vertical bar 3 x plus 1 close vertical bar plus open vertical bar 2 x minus 3 close vertical bar end cell equals 5 row cell 5 x minus 2 end cell equals 5 row cell 5 x end cell equals cell 5 plus 2 end cell row cell 5 x end cell equals 7 row x equals cell 7 over 5 end cell end table

Diperoleh nilai x equals 7 over 5, tetapi tidak memenuhi syarat x greater or equal than 3 over 2, karena 7 over 5 less than 3 over 2.

Nilai x pada x plus 4 untuk negative 1 third less or equal than x less than 3 over 2.

table attributes columnalign right center left columnspacing 2px end attributes row cell open vertical bar 3 x plus 1 close vertical bar plus open vertical bar 2 x minus 3 close vertical bar end cell equals 5 row cell x plus 4 end cell equals 5 row x equals cell 5 minus 4 end cell row x equals 1 end table

Diperoleh nilai x equals 1 memenuhi syarat negative 1 third less or equal than x less than 3 over 2.

Nilai x pada negative 5 x plus 2 untuk x less than negative 1 third.

table attributes columnalign right center left columnspacing 2px end attributes row cell open vertical bar 3 x plus 1 close vertical bar plus open vertical bar 2 x minus 3 close vertical bar end cell equals 5 row cell negative 5 x plus 2 end cell equals 5 row cell negative 5 x end cell equals cell 5 minus 2 end cell row cell negative 5 x end cell equals 3 row x equals cell fraction numerator 3 over denominator negative 5 end fraction end cell row x equals cell negative 3 over 5 end cell end table

Diperoleh nilai x equals negative 3 over 5, memenuhi syarat x less than negative 1 third.

Dengan demikian, diperoleh himpunan penyelesaiannya adalah open curly brackets negative 3 over 5 comma space 1 close curly brackets.

0

Roboguru

Nilai  yang memenuhi persamaan  adalah ...

Pembahasan Soal:

Ingat!

Definisi dari suatu nilai mutlak adalah:

open vertical bar x close vertical bar open curly brackets table row cell x comma space x greater or equal than 0 space end cell row blank row cell negative x comma space x less than 0 end cell end table close 

Maka:

 

open vertical bar 2 x plus 4 close vertical bar open curly brackets table row cell 2 x plus 4 comma space 2 x plus 4 greater or equal than 0 end cell row cell 2 x greater or equal than negative 4 end cell row cell x greater or equal than negative 2 end cell row blank row cell negative left parenthesis 2 x plus 4 right parenthesis comma space 2 x plus 4 less than 0 end cell row cell 2 x less than negative 4 end cell row cell x less than negative 2 end cell end table close 

open vertical bar 3 minus x close vertical bar open curly brackets table row cell 3 minus x comma space 3 minus x greater or equal than 0 end cell row cell negative x greater or equal than negative 3 end cell row cell x less or equal than 3 end cell row blank row cell negative left parenthesis 3 minus x right parenthesis comma space 3 minus x less than 0 end cell row cell negative x less than negative 3 end cell row cell x greater than 3 end cell end table close 

Sehingga daerah penyelesaian terbadi menjadi 3 yaitu:

Untuk setiap interval, maka:

  • Interval x less than negative 2  

table attributes columnalign right center left columnspacing 0px end attributes row cell negative open parentheses 2 x plus 4 close parentheses minus left parenthesis 3 minus x right parenthesis end cell equals cell negative 1 end cell row cell negative 2 x minus 4 minus 3 plus x end cell equals cell negative 1 end cell row cell negative x minus 7 end cell equals cell negative 1 end cell row cell negative x end cell equals cell negative 1 plus 7 end cell row cell negative x end cell equals 6 row x equals cell negative 6 end cell end table 

Karena x equals negative 6 berada pada interval x less than negative 2, maka x equals negative 6 adalah penyelesaian.

  • Interval negative 2 less or equal than x less or equal than 3

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x plus 4 minus left parenthesis 3 minus x right parenthesis end cell equals cell negative 1 end cell row cell 2 x plus 4 minus 3 plus x end cell equals cell negative 1 end cell row cell 3 x plus 1 end cell equals cell negative 1 end cell row cell 3 x end cell equals cell negative 1 minus 1 end cell row cell 3 x end cell equals cell negative 2 end cell row x equals cell negative 2 over 3 end cell end table 

Karena table attributes columnalign right center left columnspacing 0px end attributes row x equals cell negative 2 over 3 end cell end table berada pada interval negative 2 less or equal than x less or equal than 3, maka table attributes columnalign right center left columnspacing 0px end attributes row x equals cell negative 2 over 3 end cell end table merupakan penyelesaian.

  • Interval x greater than 3

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x plus 4 minus left parenthesis negative left parenthesis 3 minus x right parenthesis right parenthesis end cell equals cell negative 1 end cell row cell 2 x plus 4 plus 3 minus x end cell equals cell negative 1 end cell row cell x plus 7 end cell equals cell negative 1 end cell row x equals cell negative 1 minus 7 end cell row x equals cell negative 8 end cell end table 

Karena x equals negative 8 tidak berada pada interval x greater than 3, maka x equals negative 8 bukan penyelesaian.

Dengan demikian, nilai x yang memenuhi persamaan tersebut adalah x equals negative 6 space atau space x equals negative 2 over 3.

1

Roboguru

Tentukan himpunan penyelesaian dari persamaan berikut!

Pembahasan Soal:

Dengan menggunakan definisi nilai mutlak diperoleh:

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 2 x plus 4 close vertical bar end cell equals cell open curly brackets table attributes columnalign left end attributes row cell 2 x plus 4 comma space untuk space 2 x plus 4 greater or equal than 0 space left right double arrow x greater or equal than negative 2 end cell row cell negative open parentheses 2 x plus 4 close parentheses comma space untuk space 2 x plus 4 less than 0 space left right double arrow x less than negative 2 end cell end table close end cell row cell open vertical bar 3 minus x close vertical bar end cell equals cell open curly brackets table attributes columnalign left end attributes row cell 3 minus x comma space untuk space 3 minus x greater or equal than 0 space left right double arrow x less or equal than 3 end cell row cell negative open parentheses 3 minus x close parentheses comma space untuk space 3 minus x less than 0 space left right double arrow x greater than 3 end cell end table close end cell end table end style 

Pada interval begin mathsize 14px style x less than negative 2 end style, diperoleh penyelesaian sebagai berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell negative open parentheses 2 x plus 4 close parentheses minus open parentheses 3 minus x close parentheses end cell equals cell negative 1 end cell row cell negative 2 x minus 4 minus 3 plus x end cell equals cell negative 1 end cell row cell negative x minus 7 end cell equals cell negative 1 end cell row cell negative x end cell equals cell negative 1 plus 7 end cell row cell negative x end cell equals 6 row x equals cell negative 6 space open parentheses memenuhi close parentheses end cell end table end style 

Pada interval begin mathsize 14px style negative 2 less or equal than x less or equal than 3 end style, diperoleh penyelesaian sebagai berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 2 x plus 4 close parentheses minus open parentheses 3 minus x close parentheses end cell equals cell negative 1 end cell row cell 2 x plus 4 minus 3 plus x end cell equals cell negative 1 end cell row cell 3 x plus 1 end cell equals cell negative 1 end cell row cell 3 x end cell equals cell negative 1 minus 1 end cell row cell 3 x end cell equals cell negative 2 end cell row x equals cell negative 2 over 3 space open parentheses memenuhi close parentheses end cell end table end style 

Pada interval begin mathsize 14px style x greater than 3 end style, diperoleh penyelesaian sebagai berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 2 x plus 4 close parentheses minus open parentheses negative open parentheses 3 minus x close parentheses close parentheses end cell equals cell negative 1 end cell row cell 2 x plus 4 plus 3 minus x end cell equals cell negative 1 end cell row cell x plus 7 end cell equals cell negative 1 end cell row x equals cell negative 1 minus 7 end cell row x equals cell negative 8 space open parentheses tidak space memenuhi close parentheses end cell end table end style 

Jadi, himpunan penyelesaian dari persamaan tersebut adalah begin mathsize 14px style HP equals open curly brackets negative 6 comma space minus 2 over 3 close curly brackets end style.

0

Roboguru

Jika penyelesaian  adalah  dan  dengan , nilai dari  adalah ...

Pembahasan Soal:

open vertical bar x minus 3 close vertical bar equals open vertical bar 4 minus 3 x close vertical bar plus 1

Menurut definisi

open vertical bar x minus 3 close vertical bar equals open curly brackets table attributes columnalign left end attributes row cell x minus 3 comma space x greater or equal than 3 end cell row cell negative x plus 3 comma x less than 3 end cell end table close

open vertical bar 4 minus 3 x close vertical bar equals open curly brackets table attributes columnalign left end attributes row cell 4 minus 3 x comma space x greater or equal than 4 over 3 end cell row cell negative 4 plus 3 x comma x less than 4 over 3 end cell end table close

Sehingga diperoleh

Untuk x less than 4 over 3, maka

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar x minus 3 close vertical bar end cell equals cell open vertical bar 4 minus 3 x close vertical bar plus 1 end cell row cell negative x plus 3 end cell equals cell negative 4 plus 3 x plus 1 end cell row cell negative x plus 3 x end cell equals cell 3 minus 4 plus 1 end cell row cell 2 x end cell equals 0 row x equals 0 end table

Untuk  4 over 3 less or equal than x less than 3, maka

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar x minus 3 close vertical bar end cell equals cell open vertical bar 4 minus 3 x close vertical bar plus 1 end cell row cell negative x plus 3 end cell equals cell 4 minus 3 x plus 1 end cell row cell 3 x minus x end cell equals cell 4 plus 1 minus 3 end cell row cell 2 x end cell equals 2 row x equals cell 1 open parentheses tidak space memenuhi close parentheses end cell end table

Untuk x greater or equal than 3

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar x minus 3 close vertical bar end cell equals cell open vertical bar 4 minus 3 x close vertical bar plus 1 end cell row cell x minus 3 end cell equals cell 4 minus 3 x plus 1 end cell row cell 3 x plus x end cell equals cell 4 plus 3 plus 1 end cell row cell 4 x end cell equals 8 row x equals 2 end table

Sehingga diperoleh HP equals open curly brackets 0 comma 2 close curly brackets, dengan x subscript 1 equals 0 dan x subscript 2 equals 4 

Nilai dari open parentheses x subscript 1 plus 2 x subscript 2 close parentheses

table attributes columnalign right center left columnspacing 0px end attributes row cell x subscript 1 plus 2 x subscript 2 end cell equals cell 0 plus 2 times 2 end cell row blank equals cell 0 plus 4 end cell row blank equals 4 end table

Jadi, jawaban yang tepat adalah C

 

0

Roboguru

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