Pertanyaan

∣ 2 x − 1 ∣ ≤ ∣ x + 4 ∣

$∣ 2 x - 1 ∣ \leq ∣ x + 4 ∣$

G. Albiah

Master Teacher

Mahasiswa/Alumni Universitas Galuh Ciamis

Jawaban terverifikasi

Jawaban

himpunan penyelesaian dari adalah .

himpunan penyelesaian dari open vertical bar 2 x minus 1 close vertical bar less or equal than open vertical bar x plus 4 close vertical bar

adalah

negative 1 less or equal than thin space x less or equal than thin space 5

Pembahasan

Ingat aturan nilai mutlak! Maka, a. atau b. Di dapatkan interval . Untuk interval Untuk interval Di dapatkan dan maka Untuk interval Di dapatkan dan . Maka . Interval penyelesaian dan maka . Jadi, himpunan penyelesaian dari adalah .

Ingat aturan nilai mutlak!

open vertical bar x close vertical bar open curly brackets table attributes columnalign left end attributes row cell negative x rightwards arrow x less or equal than 0 end cell row cell x rightwards arrow x greater than 0 end cell end table close

Maka,

a. open vertical bar 2 x minus 1 close vertical bar equals 2 x minus 1

$table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x minus 1 end cell greater or equal than cell thin space 0 end cell row cell 2 x minus 1 plus 1 end cell greater or equal than cell thin space 0 plus 1 end cell row cell 2 x end cell greater or equal than cell thin space 1 end cell row cell fraction numerator 2 x over denominator 2 end fraction end cell greater or equal than cell 1 half end cell row x greater or equal than cell 1 half end cell end table$

atau

open vertical bar 2 x minus 1 close vertical bar equals negative open parentheses 2 x minus 1 close parentheses

$table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x minus 1 end cell less than 0 row cell 2 x minus 1 plus 1 end cell less than cell 0 plus 1 end cell row cell 2 x end cell less than 1 row cell fraction numerator 2 x over denominator 2 end fraction end cell less than cell 1 half end cell row x less than cell 1 half end cell row blank blank blank end table$

b. open vertical bar x plus 4 close vertical bar equals x plus 4

table attributes columnalign right center left columnspacing 0px end attributes row cell x plus 4 end cell greater or equal than cell thin space 0 end cell row cell x plus 4 minus 4 end cell greater or equal than cell thin space 0 minus 4 end cell row x greater or equal than cell thin space minus 4 end cell end table

open vertical bar x plus 4 close vertical bar equals negative open parentheses x plus 4 close parentheses

table attributes columnalign right center left columnspacing 0px end attributes row cell x plus 4 end cell less than 0 row cell x plus 4 minus 4 end cell less than cell 0 minus 4 end cell row x less than cell negative 4 end cell end table

Di dapatkan interval x less than negative 4 comma thin space 4 less or equal than thin space x less than 1 half comma thin space x greater or equal than 1 half .

Untuk interval x less or equal than negative 4

Untuk interval negative 4 less or equal than thin space x less than 1 half

$table attributes columnalign right center left columnspacing 0px end attributes row cell negative open parentheses 2 x minus 1 close parentheses end cell less or equal than cell thin space x plus 4 end cell row cell negative 2 x plus 1 end cell less or equal than cell thin space x plus 4 end cell row cell negative 2 x plus 1 minus 1 end cell less or equal than cell thin space x plus 4 minus 1 end cell row cell negative 2 x end cell less or equal than cell thin space x plus 3 end cell row cell negative 2 x minus x end cell less or equal than cell thin space x plus 3 minus x end cell row cell negative 3 x end cell less or equal than cell thin space 3 end cell row cell open parentheses negative 3 x close parentheses open parentheses negative 1 close parentheses end cell greater or equal than cell thin space 3 open parentheses negative 1 close parentheses end cell row cell 3 x end cell greater or equal than cell thin space minus 3 end cell row cell fraction numerator 3 x over denominator 3 end fraction end cell greater or equal than cell fraction numerator negative 3 over denominator 3 end fraction end cell row x greater or equal than cell thin space minus 1 end cell row blank blank blank end table$

Di dapatkan x greater or equal than negative 1 dan negative 4 less or equal than thin space x less than 1 half maka negative 1 less or equal than thin space x less than 1 half

Untuk interval thin space x greater or equal than 1 half

Di dapatkan x less or equal than thin space 5 dan x greater or equal than 1 half . Maka 1 half less or equal than thin space x less or equal than thin space 5 .

Interval penyelesaian negative 1 less or equal than thin space x less than 1 half dan 1 half less or equal than thin space x less or equal than thin space 5 maka .

Jadi, himpunan penyelesaian dari open vertical bar 2 x minus 1 close vertical bar less or equal than open vertical bar x plus 4 close vertical bar adalah negative 1 less or equal than thin space x less or equal than thin space 5 .