Giá trị của \(a,b\) thỏa mãn \(\dfrac{1}{\left(x+1\right)\left(x-1\right)}=\dfrac{a}{x+1}+\dfrac{b}{x-1}\) là
\(a=-\dfrac{1}{2};b=-\dfrac{1}{2}.\)\(a=\dfrac{1}{2};b=\dfrac{1}{2}.\)\(a=\dfrac{1}{2};b=-\dfrac{1}{2}.\)\(a=-\dfrac{1}{2};b=\dfrac{1}{2}.\)Hướng dẫn giải:Ta có: \(\dfrac{a}{x+1}+\dfrac{b}{x-1}=\dfrac{a\left(x-1\right)+b\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left(a+b\right)x+\left(b-a\right)}{\left(x-1\right)\left(x+1\right)}\).
Để \(\dfrac{1}{\left(x+1\right)\left(x-1\right)}=\dfrac{a}{x+1}+\dfrac{b}{x-1}\)
\(\Rightarrow1=\left(a+b\right)x+\left(b-a\right)\) với mọi \(x\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=0\\b-a=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}b=-a\\b=a+1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}b=-a\\-a=a+1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}b=-a\\a=-\dfrac{1}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}b=\dfrac{1}{2}\\a=-\dfrac{1}{2}\end{matrix}\right..\)
