Nilai x yang memenuhi pertidaksamaan ∣x−1∣2<∣x+1∣1 adalah ...

Pertanyaan

Nilai $x$ yang memenuhi pertidaksamaan $fraction numerator 2 over denominator open vertical bar x minus 1 close vertical bar end fraction less than fraction numerator 1 over denominator open vertical bar x plus 1 close vertical bar end fraction$ adalah ...

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I. Sutiawan

Master Teacher

Mahasiswa/Alumni Universitas Pasundan

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah C

Pembahasan

Syarat:

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar x minus 1 close vertical bar end cell not equal to 0 row cell x minus 1 end cell not equal to 0 row x not equal to cell 0 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 x plus 1 close vertical bar end cell not equal to 0 row cell x plus 1 end cell not equal to 0 row x not equal to cell negative 1 space...... space left parenthesis 2 right parenthesis end cell end table

Maka:

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

Iriskan (1), (2) dan (3) sehingga penyelesaiannya menjadi negative 3 less than x less than negative 1 space atau space minus 1 less than x less than negative 1 third .

Jadi, jawaban yang tepat adalah C