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Jika  dan  merupakan penyelesaian persamaan mutlak , nilai dari  adalah ....

Pertanyaan

Jika begin mathsize 14px style x subscript 1 end style dan begin mathsize 14px style x subscript 2 end style merupakan penyelesaian persamaan mutlak begin mathsize 14px style 3 open vertical bar 2 x minus 1 close vertical bar equals open vertical bar 2 x minus 1 close vertical bar plus 8 end style, nilai dari begin mathsize 14px style 3 open parentheses x subscript 1 plus x subscript 2 close parentheses end style adalah ....

Pembahasan Video:

Pembahasan Soal:

Perhatikan perhitungan berikut!

begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row cell 3 open vertical bar 2 x minus 1 close vertical bar end cell equals cell open vertical bar 2 x minus 1 close vertical bar plus 8 end cell row cell 3 open vertical bar 2 x minus 1 close vertical bar end cell equals cell open vertical bar 2 x minus 1 close vertical bar plus 8 end cell row cell 3 open vertical bar 2 x minus 1 close vertical bar minus open vertical bar 2 x minus 1 close vertical bar end cell equals 8 row cell 2 open vertical bar 2 x minus 1 close vertical bar end cell equals 8 row cell open vertical bar 2 x minus 1 close vertical bar end cell equals 4 row cell table attributes columnalign right center left columnspacing 2px end attributes row cell 2 x subscript 1 minus 1 end cell equals 4 row cell 2 x subscript 1 end cell equals 5 row cell x subscript 1 end cell equals cell 5 over 2 end cell end table end cell dan cell table attributes columnalign right center left columnspacing 2px end attributes row cell 2 x subscript 2 minus 1 end cell equals cell negative 4 end cell row cell 2 x subscript 2 end cell equals cell negative 3 end cell row cell x subscript 2 end cell equals cell negative 3 over 2 end cell end table end cell end table end style

Setelah diperoleh penyelesaian begin mathsize 14px style x subscript 1 end style dan begin mathsize 14px style x subscript 2 end style substitusikan ke begin mathsize 14px style 3 open parentheses x subscript 1 plus x subscript 2 close parentheses end style, diperoleh:

begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row cell 3 open parentheses x subscript 1 plus x subscript 2 close parentheses end cell equals cell 3 open parentheses 5 over 2 plus open parentheses negative 3 over 2 close parentheses close parentheses end cell row blank equals cell 3 open parentheses 2 over 2 close parentheses end cell row blank equals 3 end table end style 

Jadi, jawaban yang tepat adalah C.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

F. Freelancer7

Terakhir diupdate 03 Mei 2021

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Himpunan penyelesaian:   adalah ...

Pembahasan Soal:

Jika open vertical bar f open parentheses x close parentheses close vertical bar plus-or-minus open vertical bar g open parentheses x close parentheses close vertical bar equals c, maka berlaku:

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar f open parentheses x close parentheses close vertical bar end cell equals cell open curly brackets table attributes columnalign left end attributes row cell f open parentheses x close parentheses greater or equal than 0 end cell row cell negative f open parentheses x close parentheses less than 0 end cell end table close end cell row cell open vertical bar g open parentheses x close parentheses close vertical bar end cell equals cell open curly brackets table attributes columnalign left end attributes row cell g open parentheses x close parentheses greater or equal than 0 end cell row cell negative g open parentheses x close parentheses less than 0 end cell end table close end cell end table 

Diketahui open vertical bar 2 x plus 4 close vertical bar minus open vertical bar 3 x minus 1 close vertical bar equals negative 1, maka:

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 greater or equal than 0 end cell row cell negative open parentheses 2 x plus 4 close parentheses less than 0 end cell end table close end cell row cell 2 x plus 4 end cell greater or equal than 0 row cell 2 x end cell greater or equal than cell negative 4 end cell row x greater or equal than cell fraction numerator negative 4 over denominator 2 end fraction end cell row bold italic x bold greater or equal than cell bold minus bold 2 end cell row cell negative open parentheses 2 x plus 4 close parentheses end cell less than 0 row cell negative 2 x minus 4 end cell less than 0 row cell negative 2 x end cell less than 4 row x less than cell fraction numerator 4 over denominator negative 2 end fraction end cell row bold italic x bold less than cell bold minus bold 2 end cell row cell open vertical bar 3 x minus 1 close vertical bar end cell equals cell open curly brackets table attributes columnalign left end attributes row cell 3 x minus 1 greater or equal than 0 end cell row cell negative open parentheses 3 x minus 1 close parentheses less than 0 end cell end table close end cell row cell 3 x minus 1 end cell greater or equal than 0 row cell 3 x end cell greater or equal than 1 row bold italic x bold greater or equal than cell bold 1 over bold 3 end cell row cell negative open parentheses 3 x minus 1 close parentheses end cell less than 0 row cell negative 3 x plus 1 end cell less than 0 row cell negative 3 x end cell less than cell negative 1 end cell row x less than cell fraction numerator negative 1 over denominator negative 3 end fraction end cell row bold italic x bold less than cell bold 1 over bold 3 end cell end table 

Gambarkan 4 nilai x pada garis bilangan.



 

Berdasarkan gambar, terdapat 3 daerah penyelesaian yaitu:

  • Untuk x less than negative 2 (Hijau)

Pada daerah ini berlaku pertidaksamaan:

negative 3 x plus 1 less than 0 space minus 2 x minus 4 less than 0 

Maka,

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 2 x plus 4 close vertical bar minus open vertical bar 3 x minus 1 close vertical bar end cell equals cell negative 1 end cell row cell negative 2 x minus 4 minus open parentheses negative 3 x plus 1 close parentheses end cell equals cell negative 1 end cell row cell negative 2 x minus 4 plus 3 x minus 1 end cell equals cell negative 1 end cell row cell x minus 5 end cell equals cell negative 1 end cell row x equals cell 5 minus 1 end cell row x equals cell 4 space open parentheses tidak space memenuhi space untuk space x less than negative 2 close parentheses end cell end table 

  • Untuk negative 2 less or equal than x less or equal than 1 third (Putih)

Pada daerah ini berlaku pertidaksamaan:

table attributes columnalign right center left columnspacing 0px end attributes row cell negative 3 x plus 1 end cell less than cell 0 space end cell row cell 2 x plus 4 end cell greater or equal than 0 end table 

Maka,

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 2 x plus 4 close vertical bar minus open vertical bar 3 x minus 1 close vertical bar end cell equals cell negative 1 end cell row cell 2 x plus 4 minus open parentheses negative 3 x plus 1 close parentheses end cell equals cell negative 1 end cell row cell 2 x plus 4 plus 3 x minus 1 end cell equals cell negative 1 end cell row cell 5 x plus 3 end cell equals cell negative 1 end cell row cell 5 x end cell equals cell negative 1 minus 3 end cell row x equals cell negative 4 over 5 space open parentheses memenuhi space untuk space minus 2 less or equal than x less or equal than 1 third close parentheses end cell end table 

  • Untuk table attributes columnalign right center left columnspacing 0px end attributes row blank blank x end table table attributes columnalign right center left columnspacing 0px end attributes row blank greater than blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 1 third end cell end table (Kuning)

Pada daerah ini berlaku pertidaksamaan:

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 x minus 1 end cell greater or equal than cell 0 space end cell row cell 2 x plus 4 end cell greater or equal than 0 end table 

Maka,

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 2 x plus 4 close vertical bar minus open vertical bar 3 x minus 1 close vertical bar end cell equals cell negative 1 end cell row cell 2 x plus 4 minus open parentheses 3 x minus 1 close parentheses end cell equals cell negative 1 end cell row cell 2 x plus 4 minus 3 x plus 1 end cell equals cell negative 1 end cell row cell negative x plus 5 end cell equals cell negative 1 end cell row cell negative x end cell equals cell negative 1 minus 5 end cell row cell negative x end cell equals cell negative 6 end cell row x equals cell 6 space open parentheses memenuhi space untuk space x greater than 1 third close parentheses end cell end table 

Sehingga, himpunan penyelesaian dari open vertical bar 2 x plus 4 close vertical bar minus open vertical bar 3 x minus 1 close vertical bar equals negative 1 adalah open curly brackets negative 4 over 5 comma space 6 close curly brackets.

Jadi, tidak ada jawaban yang tepat.

0

Roboguru

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

open vertical bar x plus 5 close vertical bar open curly brackets table row cell x plus 5 comma space x plus 5 greater or equal than 0 space end cell row cell x greater or equal than negative 5 end cell row blank row cell negative left parenthesis x plus 5 right parenthesis comma space x plus 5 less than 0 end cell row cell x less than negative 5 end cell end table close 

Sehingga daerah penyelesaian terbagi menjadi 3 yaitu:

  • Untuk x less than negative 5 

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

Sehingga untuk x less than negative 2 tidak terdapat penyelesaian.

  • Untuk negative 5 less or equal than x less than negative 2 

table attributes columnalign right center left columnspacing 0px end attributes row cell negative open parentheses x plus 2 close parentheses minus open parentheses x plus 5 close parentheses minus 5 end cell equals 0 row cell negative x minus 2 minus x minus 5 minus 5 end cell equals 0 row cell negative 2 x minus 12 end cell equals 0 row cell negative 2 x end cell equals 12 row x equals cell fraction numerator 12 over denominator negative 2 end fraction end cell row x equals cell negative 6 end cell end table 

negative 6  tidak berada pada interval negative 5 less or equal than x less than negative 2, sehingga untuk negative 5 less or equal than x less than negative 2 tidak terdapat penyelesaian.

  • Untuk x greater or equal than negative 2

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

Sehingga untuk x greater or equal than negative 2 tidak terdapat penyelesaian.

Dari ketiga interval yang diuji, tidak memiliki himpunan penyelesaian atau dilambangkan dengan open curly brackets blank close curly brackets.

Oleh karena itu, jawaban yang tepat adalah C.

3

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

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.

0

Roboguru

Tentukan himpunan penyelesaian dari persamaan berikut .

Pembahasan Soal:

Sesuai definisi mutlak:

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 space jika space x greater or equal than 3 end cell row cell 3 minus x comma space space jika space x less than 3 end cell end table close

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 comma space space jika space x greater or equal than 4 end cell row cell 3 minus x comma space space jika space x less than 4 end cell end table close

Untuk straight x blank 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 3 minus x plus 8 minus 2 x end cell equals 5 row cell negative 3 x plus 11 end cell equals 5 row cell negative 3 x end cell equals cell negative 6 end cell row x equals cell 2 space left parenthesis memenuhi space x less than 3 right parenthesis end cell end table

Untuk 3 less or equal than straight 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 open parentheses x minus 3 close parentheses plus 8 minus 2 x end cell equals 5 row cell negative x plus 5 end cell equals 5 row x equals cell 0 space left parenthesis tidak space memenuhi space 3 less or equal than x less than 4 right parenthesis end cell end table

Untuk straight 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 open parentheses x minus 3 close parentheses plus open parentheses 2 x minus 8 close parentheses end cell equals 5 row cell 3 x minus 11 end cell equals 5 row cell 3 x end cell equals 16 row x equals cell 16 over 3 space left parenthesis memenuhi space x greater or equal than 4 right parenthesis end cell end table

Jadi, himpunan penyelesaian dari persamaan berikut open vertical bar x minus 3 close vertical bar plus open vertical bar 2 x minus 8 close vertical bar equals 5 adalah open curly brackets 2 comma 16 over 3 close curly brackets.

1

Roboguru

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