ây za :-)

Liên hợp nha e: \(\sqrt{a}-\sqrt{b}=\dfrac{a-b}{\sqrt{a}+\sqrt{b}}\)

Đk: \(x\ge5\)

Pt\(\Leftrightarrow\sqrt{x}+\sqrt{x-5}+\sqrt{x+7}-9=0\)

\(\Leftrightarrow\left(\sqrt{x}-3\right)+\left(\sqrt{x-5}-2\right)+\left(\sqrt{x+7}-4\right)=0\)

\(\Leftrightarrow\dfrac{x-9}{\sqrt{x}+3}+\dfrac{x-9}{\sqrt{x-5}+2}+\dfrac{x-9}{\sqrt{x+7}+4}=0\)

\(\Leftrightarrow\left(x-9\right)\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x-5}+2}+\dfrac{1}{\sqrt{x+7}+4}\right)=0\)

\(\Leftrightarrow x=9\)(tm) (vì \(\dfrac{1}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x-5}+2}+\dfrac{1}{\sqrt{x+7}+4}>0;\forall x\ge5\))

Vậy x=9

à ừ, quên :D