Pertanyaan

Selesaikan dan tuliskan himpunan penyelesaiannya dari PtNM berikut. b. ∣ ∣ 3 + x 6 − 5 x ∣ ∣ ≤ 2 1

Selesaikan dan tuliskan himpunan penyelesaiannya dari PtNM berikut.

b. $∣ ∣ \frac{6 - 5 x}{3 + x} ∣ ∣ \leq \frac{1}{2}$

I. Sutiawan

Master Teacher

Mahasiswa/Alumni Universitas Pasundan

Jawaban terverifikasi

Jawaban

himpunan penyelesaian pertidaksamaan adalah

himpunan penyelesaian pertidaksamaan $table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar fraction numerator 6 minus 5 x over denominator 3 plus x end fraction close vertical bar end cell less or equal than cell 1 half end cell end table$ adalah

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open curly brackets right enclose x 9 over 11 less or equal than x less or equal than 5 over 3 close curly brackets end cell end table

Pembahasan

Syarat: Iriskan (1) dan (2) sehingga penyelesaian menjadi . Jadi, himpunan penyelesaian pertidaksamaan adalah

Syarat:

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 plus x end cell not equal to 0 row x not equal to cell negative 3 space... space left parenthesis 1 right parenthesis end cell end table

$table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar fraction numerator 6 minus 5 x over denominator 3 plus x end fraction close vertical bar end cell less or equal than cell 1 half end cell row cell fraction numerator open vertical bar 6 minus 5 x close vertical bar over denominator open vertical bar 3 plus x close vertical bar end fraction end cell less or equal than cell 1 half end cell row cell 2 cross times open vertical bar 6 minus 5 x close vertical bar end cell less or equal than cell open vertical bar 3 plus x close vertical bar end cell row cell open vertical bar 12 minus 10 x close vertical bar end cell less or equal than cell open vertical bar 3 plus x close vertical bar end cell row cell left parenthesis left parenthesis 12 minus 10 x right parenthesis plus left parenthesis 3 plus x right parenthesis right parenthesis left parenthesis left parenthesis 12 minus 10 x minus left parenthesis 3 plus x right parenthesis right parenthesis end cell less or equal than 0 row cell left parenthesis 12 minus 10 x plus 3 plus x right parenthesis left parenthesis 12 minus 10 x minus 3 minus x right parenthesis end cell less or equal than 0 row cell left parenthesis negative 9 x plus 15 right parenthesis left parenthesis negative 11 x plus 9 right parenthesis end cell less or equal than 0 row cell 9 over 11 end cell less or equal than cell x less or equal than 15 over 9 end cell row cell 9 over 11 end cell less or equal than cell x less or equal than 5 over 3 space... space left parenthesis 2 right parenthesis end cell end table$

Iriskan (1) dan (2) sehingga penyelesaian menjadi table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 9 over 11 end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank less or equal than blank end table table attributes columnalign right center left columnspacing 0px end attributes row x less or equal than cell 5 over 3 end cell end table .

Jadi, himpunan penyelesaian pertidaksamaan $table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar fraction numerator 6 minus 5 x over denominator 3 plus x end fraction close vertical bar end cell less or equal than cell 1 half end cell end table$ adalah