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Himpunan penyelesaian dari 2 ∣ x + 3 ∣ ≥ ∣ x − 2 ∣ adalah...

Himpunan penyelesaian dari $2 ∣ x + 3 ∣ \geq ∣ x - 2 ∣$ adalah...

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Jawaban terverifikasi

Jawaban

himpunan penyelesaian pertidaksamaan di atas adalah .

himpunan penyelesaian pertidaksamaan di atas adalah $table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open curly brackets x vertical line minus 8 less or equal than x less or equal than fraction numerator negative 4 over denominator 3 end fraction comma x element of R close curly brackets end cell end table$ .

Pembahasan

Jadi, himpunan penyelesaian pertidaksamaan di atas adalah .

$table attributes columnalign right center left columnspacing 0px end attributes row cell 2 open vertical bar x plus 3 close vertical bar end cell greater or equal than cell open vertical bar x minus 2 close vertical bar end cell row cell 2 open parentheses x plus 3 close parentheses end cell greater or equal than cell open parentheses x minus 2 close parentheses end cell row cell 2 x plus 6 end cell greater or equal than cell x minus 2 end cell row cell 2 x minus x end cell greater or equal than cell negative 2 minus 6 end cell row x greater or equal than cell negative 8 end cell row cell 2 open vertical bar x plus 3 close vertical bar end cell less or equal than cell negative open vertical bar x minus 2 close vertical bar end cell row cell 2 open parentheses x plus 3 close parentheses end cell less or equal than cell negative open parentheses x minus 2 close parentheses end cell row cell 2 x plus 6 end cell less or equal than cell negative x plus 2 end cell row cell 2 x plus x end cell less or equal than cell 2 minus 6 end cell row cell 3 x end cell less or equal than cell negative 4 end cell row x less or equal than cell fraction numerator negative 4 over denominator 3 end fraction end cell row cell H P end cell equals cell open curly brackets x vertical line minus 8 less or equal than x less or equal than fraction numerator negative 4 over denominator 3 end fraction comma x element of R close curly brackets end cell end table$

Jadi, himpunan penyelesaian pertidaksamaan di atas adalah $table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open curly brackets x vertical line minus 8 less or equal than x less or equal than fraction numerator negative 4 over denominator 3 end fraction comma x element of R close curly brackets end cell end table$ .