mk cũng biết x=-4 rồi ==. mk cần biết cách làm cơ ^^

\(\Leftrightarrow\sqrt{x\left(x+4\right)}=\sqrt{\frac{\left(x+4\right)\left(x-4\right)}{2}};dkxđ;x\le-4;x\ge4\)

\(\Leftrightarrow x\left(x+4\right)=\frac{\left(x+4\right)\left(x-4\right)}{2}\)

\(\Leftrightarrow\left(x+4\right)\left(2x-x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)^2=0\Leftrightarrow x=-4\left(TM\right)\)

theo chị em nên bình phương 2 vế

\(\sqrt{4x-8}-2\sqrt{\dfrac{x-2}{4}}=3\left(x\ge2\right)\\ \Leftrightarrow2\sqrt{x-2}-\sqrt{x-2}=3\\ \Leftrightarrow\sqrt{x-2}=3\Leftrightarrow x-2=9\\ \Leftrightarrow x=11\left(tm\right)\)

ĐKXĐ: \(x\ge2\)

\(pt\Leftrightarrow2\sqrt{x-2}-\sqrt{x-2}=3\)

\(\Leftrightarrow\sqrt{x-2}=3\Leftrightarrow x-2=9\Leftrightarrow x=11\left(tm\right)\)

ĐKXĐ: \(3\ge x\ge5\)(vô lý)

Vậy pt vô nghiệm

<=> \(\sqrt{x^2-4x}=\sqrt{\dfrac{x^2}{2}-8}\)

<=> \(x^2-4x=\dfrac{x^2}{2}-8\)

<=>\(\dfrac{x^2}{2}-4x+8=0\)

<=> \(\left(\dfrac{x}{2}-2\right)\left(x-4\right)=0\)

<=> \(\left[{}\begin{matrix}\dfrac{x}{2}=2\\x=4\end{matrix}\right.< =>x=4\)

vậy x=4