Question

\(\sqrt{x-2}+\sqrt{10-x}=x^2-12x+40\)

Answer 1

\(ĐK:2\le x\le10\)

\(PT\Leftrightarrow\left(\sqrt{x-2}-2\right)+\left(\sqrt{10-x}-2\right)=x^2-12x+36\\ \Leftrightarrow\dfrac{x-6}{\sqrt{x-2}+2}+\dfrac{6-x}{\sqrt{10-x}+2}-\left(x-6\right)^2=0\\ \Leftrightarrow\left(x-6\right)\left(\dfrac{1}{\sqrt{x-2}+2}-\dfrac{1}{\sqrt{10-x}+2}-x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\left(tm\right)\\\dfrac{1}{\sqrt{x-2}+2}-\dfrac{1}{\sqrt{10-x}+2}-x+6=0\left(1\right)\end{matrix}\right.\)

Với \(x\le10\Leftrightarrow\left(1\right)\le\dfrac{1}{2\sqrt{2}+2}-\dfrac{1}{2}-10+6< 0\Leftrightarrow x\in\varnothing\)

Vậy \(x=6\)