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Penyelesaian pertidaksamaan x−3x+1​≥x−2 adalah...

Pertanyaan

Penyelesaian pertidaksamaan fraction numerator x plus 1 over denominator x minus 3 end fraction greater or equal than x minus 2 adalah...

  1. 1 less or equal than x less or equal than 3 space atau space x greater or equal than 5 

  2. 1 less or equal than x less than 3 space atau space x greater or equal than 5 

  3. x less or equal than 1 space atau space 3 less or equal than x less or equal than 5 

  4. x less or equal than 1 space atau space 3 less than x less or equal than 5 

  5. negative 5 less or equal than x less or equal than negative 1 space atau space x greater than 3  

E. Lestari

Master Teacher

Mahasiswa/Alumni Universitas Sebelas Maret

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah D.

Pembahasan

Ingat kembali pertidaksamaan rasional berikut.

  • Bentuk umum pertidaksamaan rasional adalah fraction numerator f left parenthesis x right parenthesis over denominator g left parenthesis x right parenthesis end fraction greater or equal than 0 dengan syarat g left parenthesis x right parenthesis not equal to 0.
  • left parenthesis a minus b right parenthesis left parenthesis c minus d right parenthesis equals a c minus a d minus b c plus b d

Diperoleh perhitungan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator x plus 1 over denominator x minus 3 end fraction end cell greater or equal than cell x minus 2 end cell row cell fraction numerator x plus 1 over denominator x minus 3 end fraction minus left parenthesis x minus 2 right parenthesis end cell greater or equal than 0 row cell fraction numerator x plus 1 over denominator x minus 3 end fraction minus fraction numerator left parenthesis x minus 2 right parenthesis left parenthesis x minus 3 right parenthesis over denominator x minus 3 end fraction end cell greater or equal than 0 row cell fraction numerator x plus 1 over denominator x minus 3 end fraction minus fraction numerator x squared minus 3 x minus 2 x plus 6 over denominator x minus 3 end fraction end cell greater or equal than 0 row cell fraction numerator x plus 1 over denominator x minus 3 end fraction minus fraction numerator x squared minus 5 x plus 6 over denominator x minus 3 end fraction end cell greater or equal than 0 row cell fraction numerator x plus 1 minus left parenthesis x squared minus 5 x plus 6 right parenthesis over denominator x minus 3 end fraction end cell greater or equal than 0 row cell fraction numerator x plus 1 minus x squared plus 5 x minus 6 over denominator x minus 3 end fraction end cell greater or equal than 0 row cell fraction numerator negative x squared plus 6 x minus 5 over denominator x minus 3 end fraction end cell greater or equal than 0 row cell fraction numerator left parenthesis negative x plus 1 right parenthesis left parenthesis x minus 5 right parenthesis over denominator x minus 3 end fraction end cell greater or equal than 0 row cell fraction numerator left parenthesis negative x plus 1 right parenthesis left parenthesis x minus 5 right parenthesis over denominator x minus 3 end fraction end cell equals 0 row x equals cell 1 semicolon space x equals 5 semicolon space x equals 3 end cell row blank blank blank end table 

Garis bilangannya digambarkan sebagai berikut.

Jadi penyelesaian dari pertidaksamaan di atas adalah x less or equal than 1 space atau space 3 less than x less or equal than 5.

Oleh karena itu, jawaban yang benar adalah D.

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