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Diketahui kesamaan . Nilai  adalah ....

Pertanyaan

Diketahui kesamaan begin mathsize 14px style fraction numerator negative x plus 8 over denominator left parenthesis x plus 3 right parenthesis left parenthesis 2 x minus 5 right parenthesis end fraction identical to fraction numerator A over denominator x plus 3 end fraction plus fraction numerator B over denominator 2 x minus 5 end fraction end style. Nilai begin mathsize 14px style A plus B end style adalah ....

  1. undefined   

  2. begin mathsize 14px style negative 1 end style   

  3. begin mathsize 14px style 0 end style   

  4. begin mathsize 14px style 1 end style   

  5. undefined   

Pembahasan Video:

Pembahasan Soal:

Diketahui kesamaan begin mathsize 14px style fraction numerator negative x plus 8 over denominator open parentheses x plus 3 close parentheses open parentheses 2 x minus 5 close parentheses end fraction identical to fraction numerator A over denominator x plus 3 end fraction plus fraction numerator B over denominator 2 x minus 5 end fraction end style. Oleh karena itu, didapat persamaan sebagai berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator negative x plus 8 over denominator left parenthesis x plus 3 right parenthesis left parenthesis 2 x minus 5 right parenthesis end fraction end cell identical to cell fraction numerator A over denominator x plus 3 end fraction plus fraction numerator B over denominator 2 x minus 5 end fraction end cell row cell fraction numerator negative x plus 8 over denominator left parenthesis x plus 3 right parenthesis left parenthesis 2 x minus 5 right parenthesis end fraction end cell identical to cell fraction numerator A left parenthesis 2 x minus 5 right parenthesis plus B left parenthesis x plus 3 right parenthesis over denominator left parenthesis x plus 3 right parenthesis left parenthesis 2 x minus 5 right parenthesis end fraction end cell row cell negative x plus 8 end cell equals cell A left parenthesis 2 x minus 5 right parenthesis plus B left parenthesis x plus 3 right parenthesis end cell row cell negative x plus 8 end cell equals cell 2 A x minus 5 A plus B x plus 3 B end cell row cell negative x plus 8 end cell equals cell 2 A x plus B x minus 5 A plus 3 B end cell row cell negative x plus 8 end cell equals cell left parenthesis 2 A plus B right parenthesis x plus left parenthesis negative 5 A plus 3 B right parenthesis end cell end table end style 

Dari hasil di atas bisa didapat dua persamaan, yaitu:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 2 A plus B end cell equals cell negative 1 space... space left parenthesis 1 right parenthesis end cell row cell negative 5 A plus 3 B end cell equals cell 8 space... space left parenthesis 2 right parenthesis end cell end table end style 

Kemudian, kita eliminasi persamaan begin mathsize 14px style left parenthesis 1 right parenthesis end style dan begin mathsize 14px style left parenthesis 2 right parenthesis end style untuk mencari nilai begin mathsize 14px style A end style. Samakan koefisien undefined dengan mengalikan undefinedpada persamaan undefined dan mengalikan undefined pada persamaan undefined.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 2 A plus B end cell equals cell negative 1 space space open vertical bar cross times 3 close vertical bar space space 6 A plus 3 B equals negative 3 end cell row cell negative 5 A plus 3 B end cell equals cell 8 space space open vertical bar cross times 1 close vertical bar space space minus 5 A plus 3 B equals 8 end cell end table end style

Karena koefisien undefined dari kedua persamaan sudah sama, maka dapat langsung diselesaikan menggunakan operasi pengurangan untuk menghilangkan nilai undefined.

begin mathsize 14px style space space space space space space space space space space space thin space space 6 A plus 3 B equals negative 3 fraction numerator space space minus 5 A plus 3 B equals 8 over denominator space space space space space space space space space space space space space space space space space 11 space A equals negative 11 end fraction minus 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 negative 1 end style

Selanjutnya, untuk mencari nilai undefined kita substitusikan nilai undefined ke dalam persamaan undefined.

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

Akibatnya, bisa dicari nilai undefined dengan mensubstitusikan nilai undefined dan undefined.

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

Dengan demikian, nilai undefined adalah 0.

Jadi, jawaban yang tepat adalah C.

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 12 April 2021

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