\(\frac{x+a}{x-5}+\frac{x+5}{x-a}=2\) ĐKXĐ: \(x\ne5\), \(x\ne a\)
\(\Leftrightarrow\frac{\left(x+a\right)\left(x-a\right)}{\left(x-5\right)\left(x-a\right)}+\frac{\left(x+5\right)\left(x-5\right)}{\left(x-5\right)\left(x-a\right)}=2\)\(\Leftrightarrow\frac{x^2-a^2+x^2-25}{\left(x-5\right)\left(x-a\right)}=2\)
\(\Leftrightarrow\frac{2x^2-a^2-25}{\left(x-5\right)\left(x-a\right)}=2\)\(\Leftrightarrow2x^2-a^2-25=2\left(x^2-ax-5x+5a\right)\)
\(\Leftrightarrow2x^2-a^2-25=2x^2-2ax-10x+10a\)
\(\Leftrightarrow-a^2+2ax-10a+10x-25=0\)\(\Leftrightarrow a^2-2ax+10a-10x+25=0\)
\(\Leftrightarrow a^2-2\left(x-5\right)a-10x+25=0\)\(\Leftrightarrow a^2-2\left(x-5\right)a+\left(x^2-10x+25\right)-x^2=0\)
\(\Leftrightarrow a^2-2\left(x-5\right)a+\left(x-5\right)^2=x^2\)\(\Leftrightarrow\left[a-\left(x-5\right)\right]^2=x^2\)
\(\Leftrightarrow\left(a-x+5\right)^2=x^2\)
\(\Leftrightarrow\orbr{\begin{cases}a-x+5=x\\a-x+5=-x\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}a+5=2x\\a+5=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=\frac{a+5}{2}\\a=-5\end{cases}}\)(đối chiếu ĐKXĐ)
\(\Rightarrow\orbr{\begin{cases}x=\frac{a+5}{2},a\ne5\\a=-5,x\ne\pm5\end{cases}}\)
Vậy..........
