a,\(\frac{1}{x\left(x^2+1\right)}=\frac{a}{x}+\frac{bx+c}{x^2+1}\Rightarrow\frac{1}{x\left(x^2+1\right)}=\frac{\left(a+b\right)x^2+cx+a}{x\left(x^2+1\right)}\)
Dong nhat 2 phan thuc tren ta duoc:
\(\hept{\begin{cases}a+b=0\\c=0\\a=1\end{cases}\Leftrightarrow\hept{\begin{cases}b=-1\\c=0\\a=1\end{cases}}}\)
b, \(\frac{1}{x^2-4}=\frac{a}{x-2}+\frac{b}{x+2}\Rightarrow\frac{1}{x^2-4}=\frac{\left(a+b\right)x+2\left(a-b\right)}{x^2-4}\)
Dong nhat 2 phan thuc tren ta duoc:
\(\hept{\begin{cases}\left(a+b\right)x=0\\2\left(a-b\right)=1\end{cases}\Leftrightarrow\hept{\begin{cases}a+b=0\\a-b=\frac{1}{2}\end{cases}\Leftrightarrow}\hept{\begin{cases}a=\frac{1}{4}\\b=\frac{-1}{4}\end{cases}}}\)
