ĐK: $x\ne 1; x\ne -2$.

\(\dfrac{2x}{x-1}+\dfrac{3x-2}{x+2}=\dfrac{6}{\left(x-1\right)\left(x+2\right)}\\ \Rightarrow2x\left(x+2\right)+\left(3x-2\right)\left(x-1\right)=6\\ \Leftrightarrow2x^2+4x+3x^2-3x-2x+2=6\\ \Leftrightarrow5x^2-x-4=0\\ \Leftrightarrow5x^2-5x+4x-4=0\\ \Leftrightarrow5x\left(x-1\right)+4\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x+4\right)=0\)

$\Leftrightarrow x-1=0$ hoặc $5x+4=0$
$\Leftrightarrow x=1$ (loại) hoặc $x=\dfrac{-4}5$

ko biết mk mới học lớp 5 nữa

Áp dụng BĐT Cosi ta có:
$x^2+\dfrac1{x^2}\ge 2\sqrt{x^2.\dfrac1{x^2}}=2$

$y^2+\dfrac1{y^2}\ge 2\sqrt{y^2.\dfrac1{y^2}}=2$

$\Rightarrow x^2+\dfrac1{x^2}+y^2+\dfrac1{y^2}\ge 2+2=4$

Dấu '=' xảy ra $\Leftrightarrow x=\dfrac1x; y=\dfrac1y\Leftrightarrow x=y=\pm 1$