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Jika diketahui ( 3 x − 7 ) ( 2 x − 1 ) 5 x + 3 ​ = 3 x − 7 2 A ​ − 2 x − 1 B ​ Maka nilai dari 2 A − B adalah ….

Jika diketahui

Maka nilai dari  adalah ….

  1. begin mathsize 14px style negative 3 end style

  2. undefined

  3. begin mathsize 14px style negative 1 end style

  4. begin mathsize 14px style 1 end style

  5. begin mathsize 14px style 3 end style

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S. Dwi

Master Teacher

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Pembahasan

Dari soal, diperoleh Perhatikan bahwa Sehingga diperoleh Eliminasi kedua persamaan tersebut Kemudian, perhatikan jika , maka dari (i) diperoleh Sehingga didapat Maka, jawaban yang tepat adalah E.

Dari soal, diperoleh

begin mathsize 14px style fraction numerator 5 x plus 3 over denominator left parenthesis 3 x minus 7 right parenthesis left parenthesis 2 x minus 1 right parenthesis end fraction equals fraction numerator 2 A over denominator 3 x minus 7 end fraction minus fraction numerator B over denominator 2 x minus 1 end fraction end style

Perhatikan bahwa

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 5 x plus 3 over denominator open parentheses 3 x minus 7 close parentheses open parentheses 2 x minus 1 close parentheses end fraction end cell equals cell fraction numerator 2 A over denominator 3 x minus 7 end fraction minus fraction numerator B over denominator 2 x minus 1 end fraction end cell row cell fraction numerator 5 x plus 3 over denominator open parentheses 3 x minus 7 close parentheses open parentheses 2 x minus 1 close parentheses end fraction end cell equals cell fraction numerator 2 A open parentheses 2 x minus 1 close parentheses minus B left parenthesis 3 x minus 7 right parenthesis over denominator left parenthesis 3 x minus 7 right parenthesis left parenthesis 2 x minus 1 right parenthesis end fraction end cell row cell fraction numerator 5 x plus 3 over denominator open parentheses 3 x minus 7 close parentheses open parentheses 2 x minus 1 close parentheses end fraction end cell equals cell fraction numerator open parentheses 2 A times 2 x minus 2 A times 1 close parentheses minus left parenthesis B times 3 x minus B times 7 right parenthesis over denominator left parenthesis 3 x minus 7 right parenthesis left parenthesis 2 x minus 1 right parenthesis end fraction end cell row cell fraction numerator 5 x plus 3 over denominator open parentheses 3 x minus 7 close parentheses open parentheses 2 x minus 1 close parentheses end fraction end cell equals cell fraction numerator 4 A x minus 2 A minus open parentheses 3 B x minus 7 B close parentheses over denominator open parentheses 3 x minus 7 close parentheses open parentheses 2 x minus 1 close parentheses end fraction end cell row cell fraction numerator 5 x plus 3 over denominator open parentheses 3 x minus 7 close parentheses open parentheses 2 x minus 1 close parentheses end fraction end cell equals cell fraction numerator 4 A x minus 2 A minus 3 B x plus 7 B over denominator open parentheses 3 x minus 7 close parentheses open parentheses 2 x minus 1 close parentheses end fraction end cell row cell fraction numerator 5 x plus 3 over denominator open parentheses 3 x minus 7 close parentheses open parentheses 2 x minus 1 close parentheses end fraction end cell equals cell fraction numerator 4 A x minus 3 B x minus 2 A plus 7 B over denominator open parentheses 3 x minus 7 close parentheses open parentheses 2 x minus 1 close parentheses end fraction end cell row cell fraction numerator 5 x plus 3 over denominator open parentheses 3 x minus 7 close parentheses open parentheses 2 x minus 1 close parentheses end fraction end cell equals cell fraction numerator open parentheses 4 A minus 3 B close parentheses x minus 2 A plus 7 B over denominator open parentheses 3 x minus 7 close parentheses open parentheses 2 x minus 1 close parentheses end fraction end cell end table end style

Sehingga diperoleh

begin mathsize 14px style 5 equals 4 A minus 3 B horizontal ellipsis left parenthesis straight i right parenthesis 3 equals negative 2 A plus 7 B horizontal ellipsis left parenthesis ii right parenthesis end style

Eliminasi kedua persamaan tersebut

begin mathsize 14px style space space space 4 A minus 3 B equals 5 space space space space space open vertical bar cross times 7 close vertical bar space space 28 A minus 21 B equals 35 minus 2 A plus 7 B equals 3 space space space space open vertical bar cross times 3 close vertical bar space space bottom enclose negative 6 A plus 21 B equals 9 end enclose plus 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 22 A equals 44 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 space space space A equals 2 end style

Kemudian, perhatikan jika begin mathsize 14px style A equals 2 end stylemaka dari (i) diperoleh

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 4 A minus 3 B end cell equals 5 row cell 4 times 2 minus 3 B end cell equals 5 row cell 8 minus 3 B end cell equals 5 row cell negative 3 B end cell equals cell negative 3 end cell row B equals 1 end table end style

Sehingga didapat

begin mathsize 14px style 2 A minus B equals 2 times 2 minus 1 equals 3 end style

Maka, jawaban yang tepat adalah E.

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