\(VT\)
\(A=\sqrt{x-7}+\sqrt{9-x}\)
\(\Rightarrow A^2=2+2\sqrt{\left(x-7\right)\left(9-x\right)}\le2+\left(x-7\right)+\left(9-x\right)=4\)
\(\Rightarrow A\le2\)
\(VP\)
\(B=\left(x-8\right)^2+2\ge2\)
Theo đề bài , \(A=B\Rightarrow A=B=2\)
Do đó \(x-7=9-x\Leftrightarrow x=8\)
Vậy \(x=8\)
P/s tham khảo nha