ĐKXĐ: \(x< 2\)
\(\sqrt{\dfrac{6}{3-x}}-2+\sqrt{\dfrac{8}{2-x}}-4=0\)
\(\Leftrightarrow\dfrac{\dfrac{6}{3-x}-4}{\sqrt{\dfrac{6}{3-x}}+2}+\dfrac{\dfrac{8}{2-x}-16}{\sqrt{\dfrac{8}{2-x}}+4}=0\)
\(\Leftrightarrow\dfrac{4x-6}{\left(3-x\right)\left(\sqrt{\dfrac{6}{3-x}}+2\right)}+\dfrac{16x-24}{\left(2-x\right)\left(\sqrt{\dfrac{8}{2-x}}+4\right)}=0\)
\(\Leftrightarrow\left(2x-3\right)\left(\dfrac{2}{\left(3-x\right)\left(\sqrt{\dfrac{6}{3-x}}+2\right)}+\dfrac{8}{\left(2-x\right)\left(\sqrt{\dfrac{8}{2-x}}+4\right)}\right)=0\)
\(\Leftrightarrow2x-3=0\) (do \(x< 2\Rightarrow\left\{{}\begin{matrix}3-x>0\\2-x>0\end{matrix}\right.\) nên phần trong ngoặc to luôn dương)
\(\Rightarrow x=\dfrac{3}{2}\)
Vậy pt có nghiệm duy nhất \(x=\dfrac{3}{2}\)
