Himpunan penyelesaian x2−x+6x2+x−6≤0, untuk x∈R adalah ...

Pertanyaan

Himpunan penyelesaian $fraction numerator x squared plus x minus 6 over denominator x squared minus x plus 6 end fraction less or equal than 0$ , untuk $x\in\mathbb{R}$ adalah ...

S. Ayu

Master Teacher

Mahasiswa/Alumni Universitas Muhammadiyah Prof. DR. Hamka

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah C.

Pembahasan

Bentuk umum pertidaksamaan rasional:

$fraction numerator f open parentheses x close parentheses over denominator g open parentheses x close parentheses end fraction less or equal than 0 semicolon space g open parentheses x close parentheses not equal to 0$

Pada $fraction numerator x squared plus x minus 6 over denominator x squared minus x plus 6 end fraction less or equal than 0$ , diperoleh:

$table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator x squared plus x minus 6 over denominator x squared minus x plus 6 end fraction end cell less or equal than 0 row cell fraction numerator x squared plus 3 x minus 2 x minus 6 over denominator x squared minus x plus 6 end fraction end cell less or equal than 0 row cell fraction numerator x open parentheses x plus 3 close parentheses minus 2 open parentheses x plus 3 close parentheses over denominator x squared minus x plus 6 end fraction end cell less or equal than 0 row cell fraction numerator open parentheses x minus 2 close parentheses open parentheses x plus 3 close parentheses over denominator x squared minus x plus 6 end fraction end cell less or equal than 0 end table$

Perhatikan bahwa pada penyebutnya tidak dapat dilakukan pemfaktoran, maka abaikan syarat pertidaksamaan.

Pembuat nol:

table attributes columnalign right center left columnspacing 0px end attributes row cell x minus 2 end cell equals 0 row x equals 2 row blank blank blank row cell x plus 3 end cell equals 0 row x equals cell negative 3 end cell end table

Uji titik:

Untuk interval x greater than 2 , ambil x equals 3 .

Untuk interval negative 3 less than x less than 2 , ambil x equals 0 .

$table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator x squared plus x minus 6 over denominator x squared minus x plus 6 end fraction end cell equals cell fraction numerator 0 squared plus 0 minus 6 over denominator 0 squared minus 0 plus 6 end fraction end cell row blank equals cell fraction numerator 0 plus 0 minus 6 over denominator 0 minus 0 plus 6 end fraction end cell row blank equals cell fraction numerator negative 6 over denominator 6 end fraction end cell row blank equals cell negative 1 space open parentheses minus close parentheses end cell end table$

Untuk interval x less than negative 3 , ambil x equals negative 4 .

$table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator x squared plus x minus 6 over denominator x squared minus x plus 6 end fraction end cell equals cell fraction numerator open parentheses negative 4 close parentheses squared plus open parentheses negative 4 close parentheses minus 6 over denominator open parentheses negative 4 close parentheses squared minus open parentheses negative 4 close parentheses plus 6 end fraction end cell row blank equals cell fraction numerator 16 minus 4 minus 6 over denominator 16 plus 4 plus 6 end fraction end cell row blank equals cell 6 over 26 end cell row blank equals cell 3 over 13 space open parentheses plus close parentheses end cell end table$

Garis bilangan:

Karena pertidaksamaan bertanda " less or equal than ", maka daerah penyelesaian berada pada interval yang bertanda open parentheses minus close parentheses atau bernilai negatif.

Himpunan penyelesaian $fraction numerator x squared plus x minus 6 over denominator x squared minus x plus 6 end fraction less or equal than 0$ , untuk x element of straight real numbers adalah open curly brackets right enclose x space end enclose space minus 3 less or equal than x less or equal than 2 close curly brackets .

Jadi, jawaban yang tepat adalah C.