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Tentukan himpunan penyelesaian dari persamaan nilai mutlak berikut! ∣x−6∣+∣3+2x∣=20

Pertanyaan

Tentukan himpunan penyelesaian dari persamaan nilai mutlak berikut!

open vertical bar x minus 6 close vertical bar plus open vertical bar 3 plus 2 x close vertical bar equals 20

I. Roy

Master Teacher

Mahasiswa/Alumni Universitas Negeri Surabaya

Jawaban terverifikasi

Jawaban

nilai x yang memenuhi adalah open curly brackets negative 29 comma negative 5 2 over 3 comma 7 2 over 3 comma 11 close curly brackets 

Pembahasan

Cari nilai x dengan kemungkinan-kemungkinan berikut.

  • Jika x minus 6 greater or equal than 0 left right double arrow x greater or equal than 6, maka open vertical bar x minus 6 close vertical bar equals x minus 6
    Jika 3 plus 2 x greater or equal than 0 left right double arrow x greater or equal than negative 3 over 2 comma maka open vertical bar 3 plus 2 x close vertical bar equals 3 plus 2 x

          Sehingga diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell x minus 6 plus 3 plus 2 x end cell equals 20 row cell 3 x minus 3 end cell equals 20 row cell 3 x end cell equals 23 row x equals cell 23 over 3 end cell row blank equals cell 7 2 over 3 end cell end table

  • Jika x minus 6 greater or equal than 0 left right double arrow x greater or equal than 6 , maka open vertical bar x minus 6 close vertical bar equals x minus 6

           JIka 3 plus 2 x less than 0 left right double arrow x less than negative 3 over 2 comma maka open vertical bar 3 plus 2 x close vertical bar equals negative 3 minus 2 x

           Sehingga diperoleh 

table attributes columnalign right center left columnspacing 0px end attributes row cell x minus 6 minus 3 minus 2 x end cell equals 20 row cell negative x minus 9 end cell equals 20 row cell negative x end cell equals 29 row x equals cell negative 29 end cell end table

  • Jika x minus 6 less than 0 left right double arrow x less than 6 , maka open vertical bar x minus 6 close vertical bar equals negative x plus 6

          JIka 3 plus 2 x less than 0 left right double arrow x less than negative 3 over 2 comma maka open vertical bar 3 plus 2 x close vertical bar equals negative 3 minus 2 x

         Sehingga diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell negative x plus 6 minus 3 minus 2 x end cell equals 20 row cell negative 3 x plus 3 end cell equals 20 row cell 3 x end cell equals cell 3 minus 20 end cell row x equals cell fraction numerator negative 17 over denominator 3 end fraction end cell row blank equals cell negative 5 2 over 3 end cell end table

  • Jika x minus 6 less than 0 left right double arrow x less than 6 , maka open vertical bar x minus 6 close vertical bar equals negative x plus 6

           Jika 3 plus 2 x greater or equal than 0 left right double arrow x greater or equal than negative 3 over 2 comma maka open vertical bar 3 plus 2 x close vertical bar equals 3 plus 2 x​​​​​​​

          Sehingga diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell negative x plus 6 plus 3 plus 2 x end cell equals 20 row cell x plus 9 end cell equals 20 row x equals cell 20 minus 9 end cell row x equals 11 end table       

Jadi, nilai x yang memenuhi adalah open curly brackets negative 29 comma negative 5 2 over 3 comma 7 2 over 3 comma 11 close curly brackets 

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