a) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(F=\left(\frac{2+x}{2-x}+\frac{2-x}{2+x}+\frac{2x^2+x-11}{x^2-4}\right):\frac{x-3}{2-x}\)
\(=\left(\frac{\left(2+x\right)^2}{\left(2-x\right)\left(2+x\right)}+\frac{\left(2-x\right)^2}{\left(2+x\right)\left(2-x\right)}-\frac{2x^2+x-11}{\left(2+x\right)\left(2-x\right)}\right)\cdot\frac{2-x}{x-3}\)
\(=\frac{4+4x+x^2+4-4x+x^2-2x^2-x+11}{\left(2+x\right)\left(2-x\right)}\cdot\frac{2-x}{x-3}\)
\(=\frac{-x+19}{\left(2+x\right)}\cdot\frac{1}{x-3}=\frac{19-x}{x^2-x-6}\)
b) Ta có: F=0
⇔\(\frac{19-x}{x^2-x-6}=0\)
⇔19-x=0
hay x=19(tm)
Vậy: Khi x=19 thì F=0
c) Ta có: |x-1|=3-x
⇔\(\left[{}\begin{matrix}x-1=3-x\\x-1=x-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1-3+x=0\\x-1-x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x-4=0\\2=0\left(loại\right)\end{matrix}\right.\)\(\Leftrightarrow2x-4=0\Leftrightarrow2x=4\Leftrightarrow x=2\left(ktmĐKXĐ\right)\)
Vậy: Khi |x-1|=3-x thì F không có giá trị
