Mạn phép sửa đề \(\frac{1}{x}-\frac{1}{a}+\frac{1}{b}=\frac{1}{x-\left(a+b\right)}\)ĐKXĐ x khác 0,a+b
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x-\left(a+b\right)}=\frac{1}{a}+\frac{1}{b}\)
\(\Leftrightarrow\frac{x-\left(a+b\right)-x}{x\left(x-a-b\right)}=\frac{a+b}{ab}\)
\(\Leftrightarrow-\frac{1}{x\left(x-\left(a+b\right)\right)}=\frac{1}{ab}\Leftrightarrow x\left(x-a-b\right)=-ab\)\(\Leftrightarrow x\left(x-a\right)-b\left(x-a\right)=0\Leftrightarrow\left(x-b\right)\left(x-a\right)=0\)
Để x=a là nghiệm thì \(\left\{{}\begin{matrix}x=a\ne0\\x=a\ne a+b\end{matrix}\right.\)
Để x=b là nghiệm thì \(\left\{{}\begin{matrix}x=b\ne0\\x=b\ne a+b\end{matrix}\right.\)
