a: ĐKXĐ: \(x\notin\left\{0;2\right\}\)
\(M=\dfrac{1}{x^2-2x}\left(\dfrac{x^2+4}{4}-4\right)+1\)
\(=\dfrac{1}{x\left(x-2\right)}\cdot\dfrac{x^2+4-4x}{4}+1\)
\(=\dfrac{1}{x\left(x-2\right)}\cdot\dfrac{\left(x-2\right)^2}{4}+1=\dfrac{x-2}{4x}+1=\dfrac{x-2+4x}{4x}=\dfrac{5x-2}{4x}\)
b: |4-x|=2
=>\(\left[{}\begin{matrix}4-x=2\\4-x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(loại\right)\\x=6\left(nhận\right)\end{matrix}\right.\)
Khi x=6 thì \(M=\dfrac{5\cdot6-2}{4\cdot6}=\dfrac{30-2}{24}=\dfrac{28}{24}=\dfrac{7}{6}\)
c: \(M=\dfrac{5x-2}{4x}\)
\(=\dfrac{1}{4}\cdot\dfrac{20x-8}{4x}=\dfrac{1}{4}\left(5-\dfrac{8}{4x}\right)=\dfrac{1}{4}\left(5-\dfrac{2}{x}\right)\)
Để M max thì \(5-\dfrac{2}{x}\) max
=>x=-1(nhận)
