a) Ta có: \(x^2-4x+5=\left(x^2-4x+4\right)+1=\left(x-2\right)^2+1>0\)
\(\dfrac{2x^2-1}{x^2-4x+5}\le0\Leftrightarrow2x^2-1\le0\Leftrightarrow x^2\le\dfrac{1}{2}\Leftrightarrow-\dfrac{1}{\sqrt{2}}\le x\le\dfrac{1}{\sqrt{2}}\)
b) ĐK: \(x^2-10x+25\ne0\Leftrightarrow\left(x-5\right)^2\ne0\Leftrightarrow x-5\ne0\Leftrightarrow x\ne5\)
\(\dfrac{2x^2-50}{x^2-10x+25}=\dfrac{2\left(x-5\right)\left(x+5\right)}{\left(x-5\right)^2}=\dfrac{2\left(x+5\right)}{x-5}\)
\(\dfrac{2x^2-50}{x^2-10x+25}\le0\Leftrightarrow\dfrac{2\left(x+5\right)}{x-5}\le0\)
TH1: \(\left\{{}\begin{matrix}x+5\le0\\x-5>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le-5\\x>5\end{matrix}\right.\) (vô lí)
TH2: \(\left\{{}\begin{matrix}x+5\ge0\\x-5< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-5\\x< 5\end{matrix}\right.\Leftrightarrow-5\le x< 5\)
