Question

Xác định các số a, b, c sao cho: \(\dfrac{1}{\left(x+1\right)^2.\left(x+2\right)}=\dfrac{a}{x+1}+\dfrac{b}{\left(x+1\right)^2}+\dfrac{c}{x+2}\)

Answer 1

Quy đồng vế phải:

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

\(=\dfrac{ax^2+3ax+2a+bx+2b+cx^2+2cx+c}{\left(x+1\right)^2\left(x+2\right)}\)

\(=\dfrac{\left(a+c\right)x^2+\left(3a+b+2c\right)x+2a+2b+c}{\left(x+1\right)^2\left(x+2\right)}\)

Đồng nhất hệ số với tử số vế trái ta được:

\(\left\{{}\begin{matrix}a+c=0\\3a+b+2c=0\\2a+2b+c=1\end{matrix}\right.\)  \(\Rightarrow\left\{{}\begin{matrix}a=-1\\b=1\\c=1\end{matrix}\right.\)