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Tentukan nilai x yang memenuhi persamaan .

Pertanyaan

Tentukan nilai x yang memenuhi persamaan open vertical bar 2 x minus 5 close vertical bar equals 3 plus 2 open vertical bar 7 minus x close vertical bar.

Pembahasan Soal:

Diketahui open vertical bar 2 x minus 5 close vertical bar equals 3 plus 2 open vertical bar 7 minus x close vertical bar

open vertical bar 2 x minus 5 close vertical bar equals open curly brackets table attributes columnalign left end attributes row cell 2 x minus 5 comma space text JIka end text space x greater or equal than 5 over 2 end cell row cell negative left parenthesis 2 x minus 5 right parenthesis comma space text jika  end text x less than 5 over 2. end cell end table close

open vertical bar 7 minus x close vertical bar equals open curly brackets table attributes columnalign left end attributes row cell 7 minus x comma space text JIka end text space x less or equal than 7 end cell row cell negative left parenthesis 7 minus x right parenthesis comma text  jika end text space x greater than 7. end cell end table close

Sehingga diperoleh perhitungan

text untuk end text space 5 over 2 less or equal than x less or equal than 7
table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x minus 5 end cell equals cell 3 plus 2 left parenthesis 7 minus x right parenthesis end cell row cell 2 x minus 5 end cell equals cell 3 plus 14 minus 2 x end cell row cell 2 x plus 2 x end cell equals cell 3 plus 14 plus 5 end cell row cell 4 x end cell equals 22 row x equals cell 22 over 4 end cell row x equals cell 11 over 2 end cell end table

text untuk end text space x greater than 7
table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x minus 5 end cell equals cell 3 plus 2. negative left parenthesis 7 minus x right parenthesis end cell row cell 2 x minus 5 end cell equals cell 3 minus 14 plus 2 x end cell row cell 2 x minus 2 x end cell equals cell 3 minus 14 plus 5 end cell row 0 equals cell negative 6 space text (tidak memenuhi) end text end cell end table

x less or equal than 7
table attributes columnalign right center left columnspacing 0px end attributes row cell negative left parenthesis 2 x minus 5 right parenthesis end cell equals cell 3 plus 2. left parenthesis 7 minus x right parenthesis end cell row cell negative 2 x plus 5 end cell equals cell 3 plus 14 minus 2 x end cell row cell negative 2 x plus 2 x end cell equals cell 3 plus 14 minus 5 end cell row 0 equals cell 12 text  (tidak memenuhi) end text end cell end table

Dengan demikian, nilai x yang memenuhi persamaan open vertical bar 2 x minus 5 close vertical bar equals 3 plus 2 open vertical bar 7 minus x close vertical bar adalah open parentheses 11 over 2 close parentheses.

 

 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

S. Solehuzain

Mahasiswa/Alumni Universitas Negeri Semarang

Terakhir diupdate 13 Agustus 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Tentukan himpunan penyelesaiannya! 3.

Pembahasan Soal:

Ingat 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 close vertical bar

open vertical bar x close vertical bar equals x

atau

open vertical bar x close vertical bar equals negative x

b. table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar x plus 3 close vertical bar end cell equals cell x plus 3 end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell x plus 3 end cell greater or equal than cell thin space 0 end cell row cell x plus 3 minus 3 end cell greater or equal than cell thin space 0 minus 3 end cell row x greater or equal than cell thin space minus 3 end cell end table

atau

open vertical bar x plus 3 close vertical bar equals negative open parentheses x plus 3 close parentheses

table attributes columnalign right center left columnspacing 0px end attributes row cell x plus 3 end cell less than 0 row cell x plus 3 minus 3 end cell less than cell 0 minus 3 end cell row x less than cell negative 3 end cell end table

Untuk x less than negative 3,

 table attributes columnalign right center left columnspacing 0px end attributes row cell negative x minus open parentheses x plus 3 close parentheses end cell equals 5 row cell negative x minus x minus 3 end cell equals 5 row cell negative 2 x end cell equals cell 5 plus 3 end cell row cell negative 2 x end cell equals 8 row x equals cell fraction numerator 8 over denominator negative 2 end fraction end cell row x equals cell negative 4 end cell end table

Di dapatkan x less than negative 3 dan x equals negative 4.

Untuk x greater or equal than 0 dengan x equals 1,

table attributes columnalign right center left columnspacing 0px end attributes row cell x plus x plus 3 end cell equals 5 row cell 2 x plus 3 end cell equals 5 row cell 2 x plus 3 minus 3 end cell equals cell 5 minus 3 end cell row cell 2 x end cell equals 2 row cell fraction numerator 2 x over denominator 2 end fraction end cell equals cell 2 over 2 end cell row x equals 1 row blank blank blank end table

Di dapatkan x equals 1 atau x greater or equal than 0.


Jadi, himpunan penyelesaian dari open vertical bar x close vertical bar plus open vertical bar x plus 3 close vertical bar equals 5 adalah x equals 1 atau x equals negative 4.

Roboguru

Tentukan Himpunan Penyelesaian dari: 1.

Pembahasan Soal:

Ingat kembali:

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

Penyelesaian dari open vertical bar x plus 1 close vertical bar plus open vertical bar x plus 3 close vertical bar equals 6 dapat ditentukan sebagai berikut:

  • Untuk x greater or equal than 0 

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

  • Untuk x less than 0 

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar x plus 1 close vertical bar plus open vertical bar x plus 3 close vertical bar end cell equals 6 row cell open parentheses x plus 1 close parentheses plus open parentheses negative open parentheses x plus 3 close parentheses close parentheses end cell equals 6 row cell up diagonal strike x plus 1 up diagonal strike negative x end strike minus 3 end cell equals cell 6 rightwards arrow tidak space ada space penyelesaian end cell end table 

Jadi, himpunan penyelesaian dari open vertical bar x plus 1 close vertical bar plus open vertical bar x plus 3 close vertical bar equals 6 adalah open curly brackets 1 close curly brackets.

Roboguru

Seekor burung pelikan terbang di atas laut dengan ketinggian 20 meter melihat ikan pada jarak 25 meter. Agar mendapatkan ikan, burung tersebut terbang dengan menukik ke permukaan dan menyelam dengan k...

Pembahasan Soal:

begin mathsize 14px style p equals ketinggian space burung space diatas space permukaan space laut q equals pergerakan space burung space dibawah space permukaan space air space laut end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row y equals cell open vertical bar x minus p close vertical bar plus q end cell row y equals cell open vertical bar x minus 20 close vertical bar plus 5 end cell row cell Jadi space p end cell equals 20 row q equals 5 end table end style 

Roboguru

Pembahasan Soal:

Ingat definisi nilai mutlak yaitu

open vertical bar x close vertical bar equals open curly brackets table attributes columnalign left columnspacing 1.4ex end attributes row x cell comma space x greater or equal than 0 end cell row cell negative x end cell cell comma space x less than 0 end cell end table close

Maka open vertical bar 2 x minus 7 close vertical bar plus open vertical bar 5 x plus 2 close vertical bar equals 12 memiliki 4 kondisi yang pertama 2 x minus 7 greater or equal than 0 space text dan end text space 5 x plus 2 greater or equal than 0 sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 2 x minus 7 close vertical bar plus open vertical bar 5 x plus 2 close vertical bar end cell equals 12 row cell 2 x minus 7 plus 5 x plus 2 end cell equals 12 row cell 7 x minus 5 end cell equals 12 row cell 7 x end cell equals cell 12 plus 5 end cell row cell 7 x end cell equals 17 row x equals cell 17 over 7 end cell end table

Kondisi kedua 2 x minus 7 greater or equal than 0 space text dan end text space 5 x plus 2 less than 0 sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 2 x minus 7 close vertical bar plus open vertical bar 5 x plus 2 close vertical bar end cell equals 12 row cell 2 x minus 7 plus negative open parentheses 5 x plus 2 close parentheses end cell equals 12 row cell 2 x minus 7 plus negative 5 x minus 2 end cell equals 12 row cell negative 3 x minus 9 end cell equals 12 row cell negative 3 x end cell equals cell 12 plus 9 end cell row cell negative 3 x end cell equals 19 row x equals cell negative 19 over 3 end cell end table

Kondisi ketiga 2 x minus 7 less than 0 space text dan end text space 5 x plus 2 greater or equal than 0 sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 2 x minus 7 close vertical bar plus open vertical bar 5 x plus 2 close vertical bar end cell equals 12 row cell negative open parentheses 2 x minus 7 close parentheses plus 5 x plus 2 end cell equals 12 row cell negative 2 x plus 7 plus 5 x plus 2 end cell equals 12 row cell 3 x plus 9 end cell equals 12 row cell 3 x end cell equals cell 12 minus 9 end cell row cell 3 x end cell equals 3 row x equals 1 end table

Kondisi keempat 2 x minus 7 less than 0 space text dan end text space 5 x plus 2 less than 0 sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 2 x minus 7 close vertical bar plus open vertical bar 5 x plus 2 close vertical bar end cell equals 12 row cell negative open parentheses 2 x minus 7 close parentheses plus negative open parentheses 5 x plus 2 close parentheses end cell equals 12 row cell negative 2 x plus 7 plus negative 5 x minus 2 end cell equals 12 row cell negative 7 x plus 5 end cell equals 12 row cell negative 7 x end cell equals cell 12 minus 5 end cell row cell negative 7 x end cell equals 7 row x equals cell negative 1 end cell end table

Dengan demikian, hasil dari open vertical bar 2 x minus 7 close vertical bar plus open vertical bar 5 x plus 2 close vertical bar equals 12 adalah seperti pada uraian di atas.

Roboguru

Tentukan nilai  yang memenuhi .

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 comma space x greater or equal than 0 end cell row cell negative x space space comma space x less than 0 end cell end table close 

Berdasarkan definisi nilai mutlak tersebut, maka:

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

Dan

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

Sehingga terdapat tiga kasus:

1. Untuk x less than 3

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar x minus 3 close vertical bar plus open vertical bar 2 x minus 8 close vertical bar end cell equals 5 row cell left parenthesis negative x plus 3 right parenthesis plus left parenthesis negative 2 x plus 8 right parenthesis end cell equals 5 row cell negative 3 x plus 11 end cell equals 5 row cell negative 3 x end cell equals cell 5 minus 11 end cell row cell negative 3 x end cell equals cell negative 6 end cell row x equals 2 end table 

Memenuhi karena x equals 2 berada pada domain x less than 3.

2. Untuk 3 less or equal than x less than 4 

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

Tidak memenuhi karena x equals 0 tidak berada pada domain 3 less or equal than x less than 4.

3. Untuk x greater or equal than 4 

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar x minus 3 close vertical bar plus open vertical bar 2 x minus 8 close vertical bar end cell equals 5 row cell left parenthesis x minus 3 right parenthesis plus left parenthesis 2 x minus 8 right parenthesis end cell equals 5 row cell 3 x minus 11 end cell equals 5 row cell 3 x end cell equals cell 5 plus 11 end cell row cell 3 x end cell equals 16 row x equals cell 16 over 3 end cell end table 

Memenuhi karena x equals 16 over 3 berada pada domain x greater or equal than 4.

Dengan demikian, nilai x yang memenuhi open vertical bar x minus 3 close vertical bar plus open vertical bar 2 x minus 8 close vertical bar equals 5 adalah x equals 2 dan x equals 16 over 3

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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