Pertanyaan

Himpunan penyelesaian dari x − 3 x − 6 ≥ x − 1 x − 2 adalah...

Himpunan penyelesaian dari $\frac{x - 6}{x - 3} \geq \frac{x - 2}{x - 1}$ adalah...

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Habis dalam

Klaim

E. Lestari

Master Teacher

Mahasiswa/Alumni Universitas Sebelas Maret

Jawaban terverifikasi

Pembahasan

Menentukan pembuat nol , , dan Garis bilangan Jadi, penyelesaian dari pertidaksamaan tersebut adalah .

$table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator x minus 6 over denominator x minus 3 end fraction end cell greater or equal than cell fraction numerator x minus 2 over denominator x minus 1 end fraction end cell row cell fraction numerator x minus 6 over denominator x minus 3 end fraction minus fraction numerator x minus 2 over denominator x minus 1 end fraction end cell greater or equal than 0 row cell fraction numerator left parenthesis x minus 6 right parenthesis left parenthesis x minus 1 right parenthesis minus left parenthesis x minus 2 right parenthesis left parenthesis x minus 3 right parenthesis over denominator left parenthesis x minus 3 right parenthesis left parenthesis x minus 1 right parenthesis end fraction end cell greater or equal than 0 row cell fraction numerator x squared minus x minus 6 x plus 6 minus left parenthesis x squared minus 3 x minus 2 x plus 6 right parenthesis over denominator left parenthesis x minus 3 right parenthesis left parenthesis x minus 1 right parenthesis end fraction end cell greater or equal than 0 row cell fraction numerator x squared minus 7 x plus 6 minus x squared plus 5 x minus 6 over denominator left parenthesis x minus 3 right parenthesis left parenthesis x minus 1 right parenthesis end fraction end cell greater or equal than 0 row cell fraction numerator negative 2 x over denominator left parenthesis x minus 3 right parenthesis left parenthesis x minus 1 right parenthesis end fraction end cell greater or equal than 0 row cell fraction numerator negative 2 x over denominator left parenthesis x minus 3 right parenthesis left parenthesis x minus 1 right parenthesis end fraction end cell equals 0 row blank blank blank end table$

Menentukan pembuat nol

$table attributes columnalign right center left columnspacing 0px end attributes row cell negative 2 x end cell equals 0 row x equals cell fraction numerator 0 over denominator negative 2 end fraction end cell row x equals 0 end table$ , , dan table attributes columnalign right center left columnspacing 0px end attributes row cell x minus 1 end cell equals 0 row x equals 1 end table

Garis bilangan

Jadi, penyelesaian dari pertidaksamaan tersebut adalah left curly bracket x vertical line space x less or equal than 0 space atau space 1 less than x less than 3 right curly bracket .