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

Himpunan penyelesaian:

 

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 

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A. Armanda

Master Teacher

Mahasiswa/Alumni Universitas Indraprasta PGRI

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Pembahasan

Jika , maka berlaku: Diketahui , maka: Gambarkan 4 nilai pada garis bilangan. Berdasarkan gambar, terdapat 3 daerah penyelesaian yaitu: Untuk (Hijau) Pada daerah ini berlaku pertidaksamaan: Maka, Untuk (Putih) Pada daerah ini berlaku pertidaksamaan: Maka, Untuk (Kuning) Pada daerah ini berlaku pertidaksamaan: Maka, Sehingga, himpunan penyelesaian dari adalah . Jadi, tidak ada jawaban yang tepat.

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.

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