Pertanyaan

Jika diketahui Maka nilai dari A ⋅ B ⋅ C adalah ….

Jika diketahui

Maka nilai dari A ⋅ B ⋅ C adalah ….

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R. RGFLSATU

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah D.

Pembahasan

Dari soal didapat Perhatikan bahwa Sehingga diperoleh Dari (iii) diperoleh Sehingga dari (ii) didapat Dan dari (i) didapat Jumlahkan persamaan (iv) dan (v) diperoleh Maka dari (i) didapat Dan dari (iii) didapat Sehingga, didapat hasil sebagai berikut Jadi, jawaban yang tepat adalah D.

Dari soal didapat

Perhatikan bahwa

$begin mathsize 14px style fraction numerator negative x squared minus 3 x plus 11 over denominator left parenthesis x minus 2 right parenthesis left parenthesis x squared minus 6 x plus 9 right parenthesis end fraction equals fraction numerator A over denominator x minus 2 end fraction plus fraction numerator B x plus C over denominator open parentheses x minus 3 close parentheses squared end fraction fraction numerator negative x squared minus 3 x plus 11 over denominator left parenthesis x minus 2 right parenthesis left parenthesis x squared minus 6 x plus 9 right parenthesis end fraction equals fraction numerator A open parentheses x minus 3 close parentheses squared plus left parenthesis B x plus C right parenthesis left parenthesis x minus 2 right parenthesis over denominator open parentheses x minus 2 close parentheses open parentheses x minus 3 close parentheses squared end fraction fraction numerator negative x squared minus 3 x plus 11 over denominator left parenthesis x minus 2 right parenthesis left parenthesis x squared minus 6 x plus 9 right parenthesis end fraction equals fraction numerator A left parenthesis x squared minus 6 x plus 9 right parenthesis plus left parenthesis B x plus C right parenthesis left parenthesis x minus 2 right parenthesis over denominator open parentheses x minus 2 close parentheses left parenthesis x squared minus 6 x plus 9 right parenthesis end fraction fraction numerator negative x squared minus 3 x plus 11 over denominator left parenthesis x minus 2 right parenthesis left parenthesis x squared minus 6 x plus 9 right parenthesis end fraction equals fraction numerator A x squared minus 6 A x plus 9 A plus B x squared minus 2 B x plus C x minus 2 C over denominator open parentheses x minus 2 close parentheses left parenthesis x squared minus 6 x plus 9 right parenthesis end fraction fraction numerator negative x squared minus 3 x plus 11 over denominator left parenthesis x minus 2 right parenthesis left parenthesis x squared minus 6 x plus 9 right parenthesis end fraction equals fraction numerator A x squared plus B x squared minus 6 A x minus 2 B x plus C x plus 9 A minus 2 C over denominator open parentheses x minus 2 close parentheses left parenthesis x squared minus 6 x plus 9 right parenthesis end fraction fraction numerator negative x squared minus 3 x plus 11 over denominator left parenthesis x minus 2 right parenthesis left parenthesis x squared minus 6 x plus 9 right parenthesis end fraction equals fraction numerator open parentheses A plus B close parentheses x squared plus open parentheses negative 6 A minus 2 B plus C close parentheses x plus 9 A minus 2 C over denominator open parentheses x minus 2 close parentheses open parentheses x squared minus 6 x plus 9 close parentheses end fraction end style$

Sehingga diperoleh

Dari (iii) diperoleh

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row 11 equals cell 9 A minus 2 C end cell row cell 11 plus 2 C end cell equals cell 9 A end cell row cell fraction numerator 11 plus 2 C over denominator 9 end fraction end cell equals A end table end style$

Sehingga dari (ii) didapat

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell negative 3 end cell equals cell negative 6 A minus 2 B plus C end cell row cell negative 3 end cell equals cell negative 6 times open parentheses fraction numerator 11 plus 2 C over denominator 9 end fraction close parentheses minus 2 B plus C end cell row cell negative 3 end cell equals cell negative 6 over 9 times open parentheses 11 plus 2 C close parentheses minus 2 B plus C end cell row cell negative 3 end cell equals cell negative 2 over 3 times open parentheses 11 plus 2 C close parentheses minus 2 B plus C end cell row cell negative 9 end cell equals cell negative 2 times open parentheses 11 plus 2 C close parentheses minus 6 B plus 3 C end cell row cell negative 9 end cell equals cell negative 22 minus 4 C minus 6 B plus 3 C end cell row cell negative 9 plus 22 end cell equals cell negative 6 B minus C end cell row 13 equals cell negative 6 B minus C horizontal ellipsis blank left parenthesis iv right parenthesis end cell end table end style$

Dan dari (i) didapat

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell negative 1 end cell equals cell A plus B end cell row cell negative 1 end cell equals cell open parentheses fraction numerator 11 plus 2 C over denominator 9 end fraction close parentheses plus B end cell row cell negative 9 end cell equals cell open parentheses 11 plus 2 C close parentheses plus 9 B end cell row cell negative 9 minus 11 end cell equals cell 2 C plus 9 B end cell row cell negative 20 end cell equals cell 2 C plus 9 B horizontal ellipsis left parenthesis straight v right parenthesis end cell end table end style$

Jumlahkan persamaan (iv) dan (v) diperoleh

Maka dari (i) didapat

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell negative 1 end cell equals cell A plus B end cell row cell negative 1 end cell equals cell A plus left parenthesis negative 2 right parenthesis end cell row cell negative 1 plus 2 end cell equals A row 1 equals A end table end style

Dan dari (iii) didapat

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row 11 equals cell 9 A minus 2 C end cell row 11 equals cell 9 times 1 minus 2 C end cell row 11 equals cell 9 minus 2 C end cell row cell 11 minus 9 end cell equals cell negative 2 C end cell row 2 equals cell negative 2 C end cell row cell negative 1 end cell equals C end table end style

Sehingga, didapat hasil sebagai berikut

begin mathsize 14px style A times B times C equals 1 times open parentheses negative 2 close parentheses times left parenthesis negative 1 right parenthesis equals 2 end style

Jadi, jawaban yang tepat adalah D.