\(ĐKXĐ:2\le x\le10\)
Ta có : \(VT^2=x-2+10-x+2\sqrt{\left(x-2\right)\left(10-x\right)}=8+2\sqrt{\left(x-2\right)\left(10-x\right)}\)
Theo AM - GM ta có : \(2\sqrt{\left(x-2\right)\left(10-x\right)}\le x-2+10-x=8\)
\(\Rightarrow VT^2\le8+8=16\Rightarrow VT\le4=VT\)
Dấu "=" xảy ra \(x-2=10-x\Leftrightarrow2x=12\Rightarrow x=6\)(TMĐKXĐ)
Vậy nghiệm PT là 6