Question

Giải pt: \(\frac{x-a}{b}+\frac{x-b}{a}=\frac{b}{x-a}+\frac{a}{x-b}\)với a,b \(\ne\) 0; a\(\ne\)+-b

Answer 1

Theo đầu bài ta có:
\(\frac{x-a}{b}+\frac{x-b}{a}=\frac{b}{x-a}+\frac{a}{x-b}\)
\(\Rightarrow\frac{a\left(x-a\right)+b\left(x-b\right)}{ab}=\frac{b\left(x-b\right)+a\left(x-a\right)}{\left(x-a\right)\left(x-b\right)}\)
\(\Rightarrow ab=\left(x^2-xb\right)-\left(xa-ab\right)\)
\(\Rightarrow x\left(x-b-a\right)=0\)
\(\Rightarrow\hept{\begin{cases}x=0\\x-\left(a+b\right)=0\end{cases}}\Rightarrow\hept{\begin{cases}x=0\\x=a+b\end{cases}}\)