Đặt \(A=\dfrac{4x^2-4x+8}{x^2+2}\)

\(\Leftrightarrow Ax^2+2A=4x^2-4x+8\\ \Leftrightarrow x^2\left(A-4\right)+4x+2A-8=0\)

PT bậc 2 ẩn x có nghiệm nên \(\Delta'=4-\left(2A-8\right)\left(A-4\right)\ge0\)

\(\Leftrightarrow-2A^2+16A-28\ge0\\ \Leftrightarrow4-\sqrt{2}\le A\le4+\sqrt{2}\)

Vậy \(A_{min}=4-\sqrt{2}\Leftrightarrow x=-\dfrac{2}{A-4}=-\dfrac{2}{-\sqrt{2}}=\sqrt{2}\)

\(a,\left\{{}\begin{matrix}OA=OB\\OC=OD\\\widehat{AOB}\text{ chung}\end{matrix}\right.\Rightarrow\Delta AOC=\Delta BOD\left(c.g.c\right)\\ \Rightarrow AC=BD\)

Áp dụng tc dtsbn:

\(\dfrac{3x+2}{4}=\dfrac{y}{2}=\dfrac{3x-y+2}{x}=\dfrac{3x+2+y-3x+y-2}{4+2-x}=\dfrac{2y}{6-x}\\ \Rightarrow\dfrac{y}{2}=\dfrac{2y}{6-x}\Rightarrow4y=6y-xy\Rightarrow xy-2y=0\\ \Rightarrow y\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}y=0\\x=2\end{matrix}\right.\\ \text{Với }x=2\Rightarrow\dfrac{3x+2}{4}=\dfrac{2y}{6-x}\Rightarrow\dfrac{8}{4}=\dfrac{2y}{4}=\dfrac{y}{2}\Rightarrow y=4\\ \text{Với }y=0\Rightarrow\dfrac{3x+2}{4}=\dfrac{0}{6-x}=0\\ \Rightarrow3x+2=0\Rightarrow x=-\dfrac{2}{3}\)

Vậy \(\left(x;y\right)=\left(2;4\right);\left(-\dfrac{2}{3};0\right)\)

\(\Leftrightarrow\left(2x^2-3x-4\right)^2-\left(x^2-x\right)^2=0\\ \Leftrightarrow\left(2x^2-3x-4-x^2+x\right)\left(2x^2-3x-4+x^2-x\right)=0\\ \Leftrightarrow\left(x^2-2x-4\right)\left(3x^2-4x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\left(1\right)\\3x^2-4x-4=0\left(2\right)\end{matrix}\right.\\ \Delta'\left(1\right)=1+4=5\Leftrightarrow x=1\pm\sqrt{5}\\ \Delta'\left(2\right)=4+12=16\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2+4}{3}=2\\x=\dfrac{2-4}{3}=-\dfrac{2}{3}\end{matrix}\right.\)

Vậy ...