Pertanyaan

Tentukan himpunan penyelesaian dari setiap PtNM berikut. b. ∣ ∣ ∣ ∣ x − 2 2 x + 1 ∣ ∣ − 1 ∣ ∣ ≤ 2

Tentukan himpunan penyelesaian dari setiap PtNM berikut.

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

I. Sutiawan

Master Teacher

Mahasiswa/Alumni Universitas Pasundan

Jawaban terverifikasi

Jawaban

himpunan penyelsaian pertidaksamaan adalah

himpunan penyelsaian pertidaksamaan $open vertical bar open vertical bar fraction numerator 2 x plus 1 over denominator x minus 2 end fraction close vertical bar minus 1 close vertical bar less or equal than 2$ adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open curly brackets right enclose x 5 over 3 less or equal than x less or equal than 7 comma space x not equal to 2 close curly brackets end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open curly brackets right enclose x 5 over 3 less or equal than x less or equal than 7 comma space x not equal to 2 close curly brackets end cell end table

Pembahasan

Syarat: Maka: Jadi, himpunan penyelsaian pertidaksamaan adalah

Syarat:

table attributes columnalign right center left columnspacing 0px end attributes row cell x minus 2 end cell not equal to 0 row x not equal to 2 end table

Maka:

$table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar open vertical bar fraction numerator 2 x plus 1 over denominator x minus 2 end fraction close vertical bar minus 1 close vertical bar end cell less or equal than 2 row cell negative 2 end cell less or equal than cell open vertical bar fraction numerator 2 x plus 1 over denominator x minus 2 end fraction close vertical bar minus 1 less or equal than 2 end cell row cell negative 2 plus 1 end cell less or equal than cell open vertical bar fraction numerator 2 x plus 1 over denominator x minus 2 end fraction close vertical bar minus 1 plus 1 less or equal than 2 plus 1 end cell row cell negative 1 end cell less or equal than cell open vertical bar fraction numerator 2 x plus 1 over denominator x minus 2 end fraction close vertical bar less or equal than 3 end cell row cell negative 1 end cell less or equal than cell open vertical bar fraction numerator 2 x plus 1 over denominator x minus 2 end fraction close vertical bar space dan space open vertical bar fraction numerator 2 x plus 1 over denominator x minus 2 end fraction close vertical bar less or equal than 3 end cell row cell open vertical bar fraction numerator 2 x plus 1 over denominator x minus 2 end fraction close vertical bar end cell greater or equal than cell negative 1 space dan space minus 3 less or equal than open vertical bar fraction numerator 2 x plus 1 over denominator x minus 2 end fraction close vertical bar less or equal than 3 end cell end table$

$table attributes columnalign right center left columnspacing 0px end attributes row x not equal to cell 2 space dan space minus 3 less or equal than open vertical bar fraction numerator 2 x plus 1 over denominator x minus 2 end fraction close vertical bar space dan space open vertical bar fraction numerator 2 x plus 1 over denominator x minus 2 end fraction close vertical bar less or equal than 3 end cell row x not equal to cell 2 space dan space open vertical bar fraction numerator 2 x plus 1 over denominator x minus 2 end fraction close vertical bar greater or equal than negative 3 space dan space open vertical bar fraction numerator 2 x plus 1 over denominator x minus 2 end fraction close vertical bar less or equal than 3 end cell row x not equal to cell 2 space dan space x not equal to 2 space space dan space fraction numerator 2 x plus 1 over denominator x minus 2 end fraction less or equal than negative 3 space atau space fraction numerator 2 x plus 1 over denominator x minus 2 end fraction greater or equal than 3 end cell row x not equal to cell 2 space dan space fraction numerator 2 x plus 1 over denominator x minus 2 end fraction plus 3 less or equal than 0 space atau space fraction numerator 2 x plus 1 over denominator x minus 2 end fraction minus 3 greater or equal than 0 end cell row x not equal to cell 2 space dan space fraction numerator 2 x plus 1 plus 3 left parenthesis x minus 2 right parenthesis over denominator x minus 2 end fraction less or equal than 0 space atau space fraction numerator 2 x plus 1 minus 3 left parenthesis x minus 2 right parenthesis over denominator x minus 2 end fraction greater or equal than 0 end cell row x not equal to cell 2 space dan space fraction numerator 2 x plus 1 plus x minus 6 over denominator x minus 2 end fraction less or equal than 0 space atau space fraction numerator 2 x plus 1 minus 3 x plus 6 over denominator x minus 2 end fraction greater or equal than 0 end cell row x not equal to cell 2 space dan space fraction numerator 3 x minus 5 over denominator x minus 2 end fraction less or equal than 0 space atau space fraction numerator negative x plus 7 over denominator x minus 2 end fraction greater or equal than 0 end cell row x not equal to cell 2 space dan space 5 over 3 less or equal than x less or equal than 2 space atau space 2 less or equal than x less or equal than 7 space end cell row cell 5 over 3 end cell less or equal than cell x less or equal than 7 comma space x not equal to 2 end cell end table$

Jadi, himpunan penyelsaian pertidaksamaan $open vertical bar open vertical bar fraction numerator 2 x plus 1 over denominator x minus 2 end fraction close vertical bar minus 1 close vertical bar less or equal than 2$ adalah