sorry nha em không biết

 em mới lớp 4 mà

\(\left\{{}\begin{matrix}\left|x\right|\ge2\\\left|y\right|\ge2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x^2\ge4\\y^2\ge4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x^2}\le\dfrac{1}{4}\\\dfrac{1}{y^2}\le\dfrac{1}{4}\end{matrix}\right.\)

\(\left(\dfrac{x+y}{xy}\right)^2=\dfrac{\left(x+y\right)^2}{x^2y^2}\le\dfrac{2\left(x^2+y^2\right)}{x^2y^2}=\dfrac{2}{x^2}+\dfrac{2}{y^2}\le2.\dfrac{1}{4}+2.\dfrac{1}{4}=1\)

\(\Rightarrow\dfrac{x+y}{xy}\le1\)

Dấu "=" xảy ra khi \(x=y=\pm2\)

\(4x^2+8x+2=0\)\(\Leftrightarrow4x^2+8x+4-2=0\)\(\Leftrightarrow4\left(x^2+2x+1\right)-2=0\)\(\Leftrightarrow4\left(x+1\right)^2-\left(\sqrt{2}\right)^2=0\)\(\Leftrightarrow\left[2\left(x+1\right)\right]^2-\left(\sqrt{2}\right)^2=0\)\(\Leftrightarrow\left[2\left(x+1\right)+\sqrt{2}\right]\left[2\left(x+1\right)-\sqrt{2}\right]=0\)\(\Leftrightarrow\left(2x+2+\sqrt{2}\right)\left(2x+2-\sqrt{2}\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}2x+2+\sqrt{2}=0\\2x+2-\sqrt{2}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-2-\sqrt{2}}{2}\\x=\frac{-2+\sqrt{2}}{2}\end{cases}}\)

Vậy tập nghiệm của pt đã cho là \(S=\left\{\frac{-2\pm\sqrt{2}}{2}\right\}\)

ơ ơ ko lm đc 

thay 2 => 4 may ra còn lm đc

\(\dfrac{x-3}{x-2}+\dfrac{x+2}{x}=2\)

Điều kiện: \(x\ne2;x\ne0\)

\(\Leftrightarrow\dfrac{x\left(x-3\right)}{x\left(x-2\right)}+\dfrac{\left(x-2\right)\left(x+2\right)}{x\left(x-2\right)}=\dfrac{2x\left(x-2\right)}{x\left(x-2\right)}\)

\(\Rightarrow x\left(x-3\right)+\left(x-2\right)\left(x+2\right)=2x\left(x-2\right)\)

\(\Leftrightarrow x^2-3x+x^2-4=2x^2-4x\)

\(\Leftrightarrow2x^2+3x-4=2x^2-4x\)

\(\Leftrightarrow x=4\)