Pertanyaan

Nilai x yang memenuhi pertidaksamaan ∣ ∣ x − 1 2 x + 7 ∣ ∣ ≥ 1 adalah....

Nilai x yang memenuhi pertidaksamaan $∣ ∣ \frac{2 x + 7}{x - 1} ∣ ∣ \geq 1$ adalah....

M. Nasrullah

Master Teacher

Mahasiswa/Alumni Universitas Negeri Makassar

Jawaban terverifikasi

Jawaban

nilai x yang memenuhi pertidaksamaan adalah

nilai x yang memenuhi pertidaksamaan $open vertical bar fraction numerator 2 x plus 7 over denominator x minus 1 end fraction close vertical bar greater or equal than 1$ adalah x less or equal than negative 8 space atau space x greater or equal than negative 2

x less or equal than negative 8 space atau space x greater or equal than negative 2

Pembahasan

Ingat kembali sifat nilai mutlak: Sehingga diperoleh perhitungan: Sehingga garis nilangannya: Jadi, nilai x yang memenuhi pertidaksamaan adalah

Ingat kembali sifat nilai mutlak:

$fraction numerator open vertical bar f open parentheses x close parentheses close vertical bar over denominator open vertical bar g open parentheses x close parentheses close vertical bar end fraction greater or equal than a rightwards double arrow open square brackets f open parentheses x close parentheses plus a times g open parentheses x close parentheses close square brackets open square brackets f open parentheses x close parentheses plus a times g open parentheses x close parentheses close square brackets greater or equal than 0$

Sehingga diperoleh perhitungan:

$table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar fraction numerator 2 x plus 7 over denominator x minus 1 end fraction close vertical bar end cell greater or equal than 1 row cell open square brackets open parentheses 2 x plus 7 close parentheses plus 1 open parentheses x minus 1 close parentheses close square brackets open square brackets open parentheses 2 x plus 7 close parentheses minus 1 open parentheses x minus 1 close parentheses close square brackets end cell greater or equal than 0 row cell open square brackets 2 x plus 7 plus x minus 1 close square brackets open square brackets 2 x plus 7 minus x plus 1 close square brackets end cell greater or equal than 0 row cell open square brackets 3 x plus 6 close square brackets open square brackets x plus 8 close square brackets end cell greater or equal than 0 end table$

$table row cell 3 x plus 6 equals 0 end cell cell atau space end cell cell x plus 8 equals 0 end cell row cell 3 x equals negative 6 end cell blank cell x equals negative 8 end cell row cell x equals fraction numerator negative 6 over denominator 3 end fraction end cell blank blank row cell x equals negative 2 end cell blank blank end table$

Sehingga garis nilangannya:

Jadi, nilai x yang memenuhi pertidaksamaan $open vertical bar fraction numerator 2 x plus 7 over denominator x minus 1 end fraction close vertical bar greater or equal than 1$ adalah x less or equal than negative 8 space atau space x greater or equal than negative 2