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Pertanyaan

Jika diketahui ( x − 2 ) ( 3 x 2 + 4 x − 7 ) − 4 x + 6 x 2 − 3 ​ = x − 2 A ​ + 3 x 2 + 4 x − 7 B x + C ​ Maka nilai dari A + B + C adalah ….

Jika diketahui

Maka nilai dari A + B + C adalah ….

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R. RGFLSATU

Master Teacher

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jawaban yang tepat adalah A.

jawaban yang tepat adalah A.

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Pembahasan

Dari soal diperoleh Perhatikan bahwa Sehingga diperoleh Dari (iii) diperoleh Maka persamaan (ii) dapat diubah menjadi Maka didapat sistem persamaan Eliminasi kedua persamaan tersebut Sehingga dari (i) diperoleh Dan dari (iii) didapat Maka didapat Jadi, jawaban yang tepat adalah A.

Dari soal diperoleh

begin mathsize 14px style fraction numerator negative 4 x plus 6 x squared minus 3 over denominator left parenthesis x minus 2 right parenthesis left parenthesis 3 x squared plus 4 x minus 7 right parenthesis end fraction equals fraction numerator A over denominator x minus 2 end fraction plus fraction numerator B x plus C over denominator 3 x squared plus 4 x minus 7 end fraction end style

Perhatikan bahwa

begin mathsize 14px style fraction numerator 6 x squared minus 4 x minus 3 over denominator open parentheses x minus 2 close parentheses open parentheses 3 x squared plus 4 x minus 7 close parentheses end fraction equals fraction numerator A over denominator x minus 2 end fraction plus fraction numerator B x plus C over denominator 3 x squared plus 4 x minus 7 end fraction fraction numerator 6 x squared minus 4 x minus 3 over denominator open parentheses x minus 2 close parentheses open parentheses 3 x squared plus 4 x minus 7 close parentheses end fraction equals fraction numerator A open parentheses 3 x squared plus 4 x minus 7 close parentheses plus open parentheses B x plus C close parentheses open parentheses x minus 2 close parentheses over denominator open parentheses x minus 2 close parentheses open parentheses 3 x squared plus 4 x minus 7 close parentheses end fraction fraction numerator 6 x squared minus 4 x minus 3 over denominator open parentheses x minus 2 close parentheses open parentheses 3 x squared plus 4 x minus 7 close parentheses end fraction equals fraction numerator 3 A x squared plus 4 A x minus 7 A plus B x squared minus 2 B x plus C x minus 2 C over denominator open parentheses x minus 2 close parentheses open parentheses 3 x squared plus 4 x minus 7 close parentheses end fraction fraction numerator 6 x squared minus 4 x minus 3 over denominator open parentheses x minus 2 close parentheses open parentheses 3 x squared plus 4 x minus 7 close parentheses end fraction equals fraction numerator 3 A x squared plus 4 A x minus 7 A plus B x squared minus open parentheses 2 B minus C close parentheses x minus 2 C over denominator open parentheses x minus 2 close parentheses open parentheses 3 x squared plus 4 x minus 7 close parentheses end fraction fraction numerator 6 x squared minus 4 x minus 3 over denominator open parentheses x minus 2 close parentheses open parentheses 3 x squared plus 4 x minus 7 close parentheses end fraction equals fraction numerator 3 A x squared plus B x squared plus 4 A x minus open parentheses 2 B minus C close parentheses x minus 7 A minus 2 C over denominator open parentheses x minus 2 close parentheses open parentheses 3 x squared plus 4 x minus 7 close parentheses end fraction fraction numerator 6 x squared minus 4 x minus 3 over denominator open parentheses x minus 2 close parentheses open parentheses 3 x squared plus 4 x minus 7 close parentheses end fraction equals fraction numerator open parentheses 3 A plus B close parentheses x squared plus open parentheses 4 A minus 2 B plus C close parentheses x minus 7 A minus 2 C over denominator open parentheses x minus 2 close parentheses open parentheses 3 x squared plus 4 x minus 7 close parentheses end fraction end style

Sehingga diperoleh

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row 6 equals cell 3 A plus B horizontal ellipsis open parentheses straight i close parentheses end cell row cell negative 4 end cell equals cell 4 A minus 2 B plus C horizontal ellipsis open parentheses ii close parentheses end cell row cell negative 3 end cell equals cell negative 7 A minus 2 C horizontal ellipsis left parenthesis iii right parenthesis end cell end table end style

Dari (iii) diperoleh

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell negative 3 end cell equals cell negative 7 A minus 2 C end cell row cell 2 C end cell equals cell negative 7 A plus 3 end cell row C equals cell fraction numerator negative 7 A plus 3 over denominator 2 end fraction end cell end table end style

Maka persamaan (ii) dapat diubah menjadi

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell negative 4 end cell equals cell 4 A minus 2 B plus C end cell row cell negative 4 end cell equals cell 4 A minus 2 B plus open parentheses fraction numerator negative 7 A plus 3 over denominator 2 end fraction close parentheses end cell row cell negative 8 end cell equals cell 8 A minus 4 B plus negative 7 A plus 3 end cell row cell negative 8 end cell equals cell A minus 4 B plus 3 end cell row cell negative 8 minus 3 end cell equals cell A minus 4 B end cell row cell negative 11 end cell equals cell A minus 4 B horizontal ellipsis left parenthesis iv right parenthesis end cell end table end style

Maka didapat sistem persamaan

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row 6 equals cell 3 A plus B horizontal ellipsis open parentheses straight i close parentheses end cell row cell negative 11 end cell equals cell A minus 4 B horizontal ellipsis left parenthesis iv end cell end table end style

Eliminasi kedua persamaan tersebut

begin mathsize 14px style 3 A plus B equals 6 space space space space space space space space space open vertical bar cross times 1 close vertical bar space 3 A plus B equals 6 A minus 4 B equals negative 11 space space vertical line cross times 3 vertical line bottom enclose 3 A minus 12 B equals negative 33 end enclose minus space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space 13 B equals 39 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space B equals 3 end style

Sehingga dari (i) diperoleh

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row 6 equals cell 3 A plus B end cell row 6 equals cell 3 A plus 3 end cell row cell 6 minus 3 end cell equals cell 3 A end cell row 3 equals cell 3 A end cell row A equals 1 end table end style

Dan dari (iii) didapat

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell negative 3 end cell equals cell negative 7 A minus 2 C end cell row cell negative 3 end cell equals cell negative 7 times 1 minus 2 C end cell row cell negative 3 end cell equals cell negative 7 minus 2 C end cell row cell negative 3 plus 7 end cell equals cell negative 2 C end cell row 4 equals cell negative 2 C end cell row cell negative 2 end cell equals C end table end style

Maka didapat

begin mathsize 14px style A plus B plus C equals 1 plus 3 plus open parentheses negative 2 close parentheses equals 2 end style

Jadi, jawaban yang tepat adalah A.

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Diberikan persamaan x 2 + 3 x + 2 − 2 x + 7 ​ = x + 1 M ​ + x + 2 N ​ Nilai dari M − N adalah ….

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