Áp dụng Cosi có \(a^2+b^2\ge2ab\left(1\right),\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{2}{ab}\left(2\right)\)
Nhân (1) và (2) có \(\left(a^2+b^2\right)\left(\frac{1}{a^2}+\frac{1}{b^2}\right)\ge4\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{4}{a^2+b^2}=\frac{4}{10}=\frac{2}{5}\)
Vậy Min q=2/5 khi a=-b
\(\frac{8x+8-8}{\left(x-1\right)\left(x+1\right)}=\frac{8\left(x+1\right)-8}{\left(x-1\right)\left(x+1\right)}=\frac{8}{x-1}-\frac{8}{\left(x-1\right)\left(x+1\right)}\)
\(x-1< x+1\Rightarrow\left\{{}\begin{matrix}x-1=1,2,-8,-4\\x+1=-1,-2,8,4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=2,3,-7,-3\\x=-2,-3,7,3\end{matrix}\right.\Rightarrow x=-3,3}\)
