Question

Xác định a,b,c sao cho:

\(\dfrac{x^2+2x-1}{\left(x-1\right)\left(x^2+1\right)}=\dfrac{a}{x-1}+\dfrac{bx+c}{x^2+1}\)

giúp mình nha,mai thi rồi

Answer 1

Ta có: \(\dfrac{x^2+2x-1}{\left(x-1\right)\left(x^2+1\right)}\)=\(\dfrac{a}{x-1}\)+\(\dfrac{bx+c}{x^2+1}\)

<=>\(\dfrac{x^2+2x-1}{\left(x-1\right)\left(x^2+1\right)}\)=\(\dfrac{a\left(x^2+1\right)+\left(bx+c\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+1\right)}\)

=>a(x²+1)+(bx+c)(x-1)=x²+2x-1

<=>ax²+a+bx²-bx+cx-c=x²+2x-1

<=>(a+b)x²+(c-b)x-(c-a)=x²+2x-1

=>\(\left\{{}\begin{matrix}a+b=1\\c-b=2\\c-a=1\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}a+b=1\\b=c-2\\a=c-1\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}c-1+c-2=1\\b=c-2\\a=c-1\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}c=2\\b=0\\a=1\end{matrix}\right.\)

Vậy a=1,b=0,c=2