Roboguru

Diberikan persamaan  dengan a,b dan c adalah konstanta. Maka nilai  adalah .....

Pertanyaan

Diberikan persamaan fraction numerator 2 x cubed plus x plus 3 over denominator left parenthesis x squared minus 1 right parenthesis left parenthesis x plus 2 right parenthesis end fraction equals fraction numerator straight a over denominator x minus 1 end fraction plus fraction numerator straight b over denominator x plus 1 end fraction plus fraction numerator straight c over denominator x plus 2 end fraction dengan a,b dan c adalah konstanta. Maka nilai straight a plus straight b plus straight c adalah .....

Pembahasan Soal:

Diketahui :

fraction numerator 2 x cubed plus x plus 3 over denominator left parenthesis x squared minus 1 right parenthesis left parenthesis x plus 2 right parenthesis end fraction equals fraction numerator straight a over denominator x minus 1 end fraction plus fraction numerator straight b over denominator x plus 1 end fraction plus fraction numerator straight c over denominator x plus 2 end fraction dengan a, b dan c konstanta.

Ditanya :
Nilai straight a plus straight b plus straight c ?

Jawab :
Dengan menggunakan konsep kesamaan dua suku banyak untuk mencari nilai a, b dan c.

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 2 x cubed plus x plus 3 over denominator left parenthesis x squared minus 1 right parenthesis left parenthesis x plus 2 right parenthesis end fraction end cell equals cell fraction numerator straight a over denominator x minus 1 end fraction plus fraction numerator straight b over denominator x plus 1 end fraction plus fraction numerator straight c over denominator x plus 2 end fraction end cell row blank equals cell fraction numerator a left parenthesis x plus 1 right parenthesis left parenthesis x plus 2 right parenthesis plus b left parenthesis x minus 1 right parenthesis left parenthesis x plus 2 right parenthesis plus c left parenthesis x minus 1 right parenthesis left parenthesis x plus 1 right parenthesis over denominator left parenthesis x minus 1 right parenthesis left parenthesis x plus 1 right parenthesis left parenthesis x plus 2 right parenthesis end fraction end cell row blank equals cell fraction numerator a left parenthesis x squared plus 3 x plus 2 right parenthesis plus b left parenthesis x squared plus x minus 2 right parenthesis plus c left parenthesis x squared minus 1 right parenthesis over denominator left parenthesis x squared minus 1 right parenthesis left parenthesis x plus 2 right parenthesis end fraction end cell row blank equals cell fraction numerator left parenthesis a plus b plus c right parenthesis x squared plus left parenthesis 3 a plus b right parenthesis x plus 2 a minus 2 b minus c over denominator left parenthesis x squared minus 1 right parenthesis left parenthesis x plus 2 right parenthesis end fraction end cell end table

Jadi, berdasarkan kesamaan di atas, pada koefisien x squared diperoleh straight a plus straight b plus straight c equals 2.

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 02 Juni 2021

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ikuti Kami

©2021 Ruangguru. All Rights Reserved