Question

giai  phuong trinh 

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

Answer 1

ĐK: \(2\le x\le10\)

\(VP=x^2-12x+40=x^2-12x+36+4=\left(x-6\right)^2+4\ge4\forall x\)

\(VT=\sqrt{x-2}+\sqrt{10-x}\le\dfrac{x-2+10-x}{2}=4\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x-6\right)^2=0\\\sqrt{x-2}=\sqrt{10-x}\end{matrix}\right.\Leftrightarrow x=6\left(tmđk\right)\)