Pertanyaan

Nilai x yang memenuhi pertidaksamaan x + 5 ∣ x − 1 ∣ ≤ 1 adalah ...

Nilai $x$ yang memenuhi pertidaksamaan $\frac{∣ x - 1 ∣}{x + 5} \leq 1$ adalah ...

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S. Eka

Master Teacher

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah E.

Pembahasan

Ingat bahwa: Nilai mutlak memiliki sifat sedangkan bentuk akar memiliki sifat . Lakukan uji titik untuk mencari daerah penyelesaian pertidaksamaan tersebut dan substitusikan pada . Misal Misal Misal Sehingga diperoleh nilai yang memenuhi pertidaksamaan adalah . Jadi, jawaban yang tepat adalah E.

Ingat bahwa:

Nilai mutlak memiliki sifat open vertical bar a close vertical bar squared equals a squared sedangkan bentuk akar memiliki sifat open parentheses square root of a close parentheses squared equals a .

$table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator open vertical bar x minus 1 close vertical bar over denominator square root of x plus 5 end root end fraction end cell less or equal than 1 row cell open vertical bar x minus 1 close vertical bar end cell less or equal than cell 1 cross times square root of x plus 5 end root end cell row cell open vertical bar x minus 1 close vertical bar end cell less or equal than cell square root of x plus 5 end root end cell row cell open parentheses x minus 1 close parentheses squared end cell less or equal than cell open parentheses square root of x plus 5 end root close parentheses squared end cell row cell x squared minus 2 x plus 1 end cell less or equal than cell x plus 5 end cell row cell x squared minus 2 x minus x plus 1 minus 5 end cell less or equal than 0 row cell x squared minus 3 x minus 4 end cell less or equal than 0 row cell open parentheses x minus 4 close parentheses open parentheses x plus 1 close parentheses end cell less or equal than 0 row cell x minus 4 end cell less or equal than 0 row x less or equal than 4 row blank blank atau row cell x plus 1 end cell less or equal than 0 row x less or equal than cell negative 1 end cell end table$

Lakukan uji titik untuk mencari daerah penyelesaian pertidaksamaan tersebut dan substitusikan pada $fraction numerator open vertical bar x minus 1 close vertical bar over denominator square root of x plus 5 end root end fraction less or equal than 1$ .

Misal x equals negative 2

$table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator open vertical bar negative 2 minus 1 close vertical bar over denominator square root of negative 2 plus 5 end root end fraction end cell less or equal than 1 row cell fraction numerator open vertical bar negative 3 close vertical bar over denominator square root of 3 end fraction end cell less or equal than 1 row cell fraction numerator 3 over denominator square root of 3 end fraction end cell less or equal than cell 1 horizontal ellipsis open parentheses salah close parentheses end cell row blank blank blank end table$

Misal x equals 0

$table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator open vertical bar 0 minus 1 close vertical bar over denominator square root of 0 plus 5 end root end fraction end cell less or equal than 1 row cell fraction numerator open vertical bar negative 1 close vertical bar over denominator square root of 5 end fraction end cell less or equal than 1 row cell fraction numerator 1 over denominator square root of 5 end fraction end cell less or equal than cell 1 horizontal ellipsis open parentheses benar close parentheses end cell end table$

Misal x equals 5

$table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator open vertical bar 5 minus 1 close vertical bar over denominator square root of 5 plus 5 end root end fraction end cell less or equal than 1 row cell fraction numerator open vertical bar 4 close vertical bar over denominator square root of 10 end fraction end cell less or equal than 1 row cell fraction numerator 4 over denominator square root of 10 end fraction end cell less or equal than cell 1 horizontal ellipsis open parentheses salah close parentheses end cell end table$

Sehingga diperoleh nilai yang memenuhi pertidaksamaan $fraction numerator open vertical bar x minus 1 close vertical bar over denominator square root of x plus 5 end root end fraction less or equal than 1$ adalah negative 1 less or equal than x less or equal than 4 .

Jadi, jawaban yang tepat adalah E.