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Himpunan penyelesaian: ∣2x+4∣−∣3x−1∣=−1  adalah ...

Pertanyaan

Himpunan penyelesaian:

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 ...space 

  1. open curly brackets 2 over 3 comma space 6 close curly brackets 

  2. open curly brackets negative 2 over 3 comma space minus 6 close curly brackets 

  3. open curly brackets negative 2 over 3 comma space 6 close curly brackets 

  4. open curly brackets 2 over 3 comma space 3 close curly brackets 

  5. open curly brackets 2 over 3 comma space minus 3 close curly brackets 

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.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Acfreelance

Mahasiswa/Alumni Universitas Indraprasta PGRI

Terakhir diupdate 06 Oktober 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Himpunan penyelesaian persamaan ∣x+2∣−∣x+5∣−5=0 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

Tentukan nilai x yang memenuhi persamaan berikut ini! 5) ∣3x+4∣+∣3−2x∣=14

Pembahasan Soal:

Ingat bahwa:

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

Diketahui open vertical bar 3 x plus 4 close vertical bar plus open vertical bar 3 minus 2 x close vertical bar equals 14,
Untuk open vertical bar 3 x plus 4 close vertical bar:

open vertical bar 3 x plus 4 close vertical bar equals open curly brackets table attributes columnalign left end attributes row cell 3 x plus 4 comma space table row cell 3 x plus 4 greater or equal than 0 end cell row cell 3 x greater or equal than negative 4 end cell row cell x greater or equal than negative 4 over 3 end cell end table end cell row cell negative left parenthesis 3 x plus 4 right parenthesis comma space table row cell negative 3 x minus 4 less than 0 end cell row cell negative 3 x less than 4 end cell row cell x less than negative 4 over 3 end cell end table end cell end table close  

Untuk open vertical bar 3 minus 2 x close vertical bar:

open vertical bar 3 minus 2 x close vertical bar equals open curly brackets table attributes columnalign left end attributes row cell 3 minus 2 x space table row cell 3 minus 2 x greater or equal than 0 end cell row cell negative 2 x greater or equal than negative 3 end cell row cell x greater or equal than 3 over 2 end cell end table end cell row cell negative left parenthesis 3 minus 2 x right parenthesis comma space table row cell negative 3 plus 2 x less than 0 end cell row cell 2 x less than 3 end cell row cell x less than 3 over 2 end cell end table end cell end table close  

Sehingga jika dibuat garis bilangan, didapat:

 

Pertidaksamaan untuk Daerah I adalah x less than negative 4 over 3 space dan space x less than 3 over 2, berdasarkan definisi nilai mutlak diatas, diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell negative open parentheses 3 x plus 4 close parentheses minus open parentheses 3 minus 2 x close parentheses end cell equals 14 row cell negative 3 x minus 4 minus 3 plus 2 x end cell equals 14 row cell negative x minus 7 end cell equals 14 row cell negative x minus 7 plus 7 end cell equals cell 14 plus 7 end cell row cell negative x end cell equals 21 row x equals cell negative 21 rightwards arrow Terpenuhi end cell end table   

Pertidaksamaan untuk Daerah II adalah x greater or equal than negative 4 over 3 space dan space x less than 3 over 2, berdasarkan definisi nilai mutlak diatas, diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 3 x plus 4 close parentheses minus open parentheses 3 minus 2 x close parentheses end cell equals 14 row cell 3 x minus open parentheses 3 minus 2 x close parentheses end cell equals 10 row cell 3 x minus 3 plus 2 x end cell equals 10 row cell 5 x minus 3 end cell equals 10 row cell 5 x end cell equals 13 row x equals cell 13 over 5 rightwards arrow Tidak space Terpenuhi end cell end table   

Pertidaksamaan untuk Daerah III adalah x greater or equal than negative 4 over 3 space dan space x greater or equal than 3 over 2, berdasarkan definisi nilai mutlak diatas, diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 x plus 4 plus 3 minus 2 x end cell equals 14 row cell x plus 4 plus 3 end cell equals 14 row cell x plus 7 end cell equals 14 row x equals cell 7 rightwards arrow Terpenuhi end cell end table   

Dari perhitungan diatas, didapat bahwa nilai x yang memenuhi adalah x equals negative 21 space dan space x equals 7.

Jadi, nilai x dari persamaan nilai mutlak tersebut adalah x equals open curly brackets negative 21 comma space 7 close curly brackets.

0

Roboguru

Himpunan penyelesaian dari persamaan ∣2x∣+∣x−1∣=5 adalah...

Pembahasan Soal:

Ingatlah definisi mutlak berikut:

table attributes columnalign right center left columnspacing 0px end attributes row cell vertical line a vertical line end cell equals cell a comma space a greater or equal than 0 end cell row blank equals cell negative a comma space a less than 0 end cell end table

Maka,

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 2 x close vertical bar end cell equals cell 2 x comma space x greater or equal than 0 end cell row blank equals cell negative 2 x comma space x less than 0 end cell row blank blank blank row cell open vertical bar x minus 1 close vertical bar end cell equals cell x minus 1 comma space x minus 1 greater or equal than 0 end cell row blank equals cell negative left parenthesis x minus 1 right parenthesis comma space x minus 1 less than 0 end cell end table

Dari uraian tersebut, akan dibagi menjadi 3 kasus,

(i) x less than 0

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

(ii) 0 less or equal than x less than 1

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

(iii) x greater or equal than 1

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar 2 x close vertical bar plus open vertical bar x minus 1 close vertical bar end cell equals 5 row cell 2 x plus left parenthesis x minus 1 right parenthesis end cell equals 5 row cell 2 x plus x minus 1 end cell equals 5 row cell 3 x minus 1 end cell equals 5 row cell 3 x end cell equals cell 5 plus 1 end cell row cell 3 x end cell equals 6 row x equals cell 2 space left parenthesis memenuhi right parenthesis end cell end table

Jadi, jawaban yang tepat adalah E.

1

Roboguru

Himpunan penyelesaian dari persamaan ∣x+2∣+4=∣2x+1∣ adalah ...

Pembahasan Soal:

Himpunan penyelesaian dari persamaan begin mathsize 14px style open vertical bar x plus 2 close vertical bar plus 4 equals open vertical bar 2 x plus 1 close vertical bar end style adalah

Sebelum menjawab pertanyaan tersebut, kita harus mengetahui konsep persamaan nilai mutlak yaitu:

begin mathsize 14px style open vertical bar x plus 2 close vertical bar equals open curly brackets table attributes columnalign left columnspacing 1.4ex end attributes row cell x plus 2 comma end cell cell x greater or equal than negative 2 end cell row cell negative x minus 2 end cell cell x less than negative 2 end cell end table close end style 

begin mathsize 14px style open vertical bar 2 x plus 1 close vertical bar equals open curly brackets table attributes columnalign left columnspacing 1.4ex end attributes row cell 2 x plus 1 comma end cell cell x greater or equal than negative 1 half end cell row cell negative 2 x minus 1 end cell cell x less than negative 1 half end cell end table close end style 

Berarti akan ada 3 kondisi:

 begin mathsize 14px style 1. space untuk space x less than negative 2 end style

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

begin mathsize 14px style 2. space untuk space minus 2 less or equal than x less than negative 1 half end style 

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

Tidak terpenuhi untuk begin mathsize 14px style x equals negative 7 over 3 end style karena tidak terletak begin mathsize 14px style negative 2 less or equal than x less than negative 1 half end style  

begin mathsize 14px style 3. space untuk space straight x greater or equal than negative 1 half end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses x plus 2 close parentheses plus 4 end cell equals cell open parentheses 2 x plus 1 close parentheses end cell row cell x minus 2 x end cell equals cell 1 minus 2 minus 4 end cell row cell negative x end cell equals cell negative 5 end cell row x equals 5 end table end style 

 

Dengan demikian, himpunan penyelesaian dari persamaan begin mathsize 14px style open vertical bar x plus 2 close vertical bar plus 4 equals open vertical bar 2 x plus 1 close vertical bar end style adalah begin mathsize 14px style open curly brackets negative 3 comma 5 close curly brackets end style 

Jadi, tidak ada jawaban yang tepat.

1

Roboguru

Tentukan himpunan penyelesaian dari persamaan dan pertidaksaman nilai mutlak berikut. a. ∣x∣+∣x−5∣=7 td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell begin mathsize 14px style vertical line x vertical line end style begin mathsize 14px style left curly bracket table attributes columnalign left end attributes row cell x space jika space x greater or equal than 0 end cell row cell negative x space jika space x less than 0 end cell end table end style end cell row blank blank cell begin mathsize 14px style vertical line x minus 5 vertical line end style begin mathsize 14px style left curly bracket table attributes columnalign left end attributes row cell x minus 5 space jika space x greater or equal than 5 end cell row cell 5 minus x space jika space x less than 5 end cell end table end style end cell row blank blank blank row cell size 14px Kondisi size 14px space size 14px 1 size 14px space size 14px colon size 14px space size 14px x end cell size 14px less than size 14px 0 row cell size 14px minus size 14px x size 14px plus size 14px left parenthesis size 14px 5 size 14px minus size 14px x size 14px right parenthesis end cell size 14px equals size 14px 7 row cell size 14px minus size 14px 2 size 14px x size 14px plus size 14px 5 end cell size 14px equals size 14px 7 row cell size 14px minus size 14px 2 size 14px x end cell size 14px equals size 14px 2 row size 14px x size 14px equals cell size 14px minus size 14px 1 size 14px space size 14px memenuhi size 14px space size 14px syarat size 14px space size 14px kondisi size 14px space size 14px x size 14px less than size 14px 0 size 14px space end cell row blank blank blank row cell size 14px Kondisi size 14px space size 14px 2 size 14px space size 14px colon size 14px 0 end cell size 14px less or equal than cell size 14px x size 14px less than size 14px 5 end cell row cell size 14px x size 14px plus size 14px left parenthesis size 14px 5 size 14px minus size 14px x size 14px right parenthesis end cell size 14px equals size 14px 7 row size 14px 5 size 14px equals cell size 14px 7 size 14px space size 14px salah size 14px space end cell row blank blank blank row blank blank blank row cell size 14px Kondis size 14px i size 14px space size 14px 3 size 14px space size 14px colon size 14px x end cell size 14px greater or equal than size 14px 5 row cell size 14px x size 14px plus size 14px x size 14px minus size 14px 5 end cell size 14px equals size 14px 7 row cell size 14px 2 size 14px x end cell size 14px equals size 14px 12 row size 14px x size 14px equals cell size 14px 6 size 14px space size 14px memenuhi size 14px space size 14px syarat size 14px space size 14px kondisi end cell row blank blank blank row blank blank blank row blank blank blank end table 

 

Jadi himpunan yang memenuhi adalah begin mathsize 14px style open parentheses negative 1 comma 6 close parentheses end style  

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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