[tex]\sqrt{5-x}+\sqrt{x-1}[/tex] =[tex]-x^{2}+2x+1[/tex]
ĐK: $1\le x\le 5$.
Ta có:
VT$^2=5-x+x-1+2\sqrt{(5-x)(x-1)}\ge 4$.
$\Rightarrow$ VT $\ge 2$.
Dấu '=' xảy ra khi $x=5$ or $x=1$.
VP $=-x^2+2x+1=2-(x^2-2x+1)=2-(x-1)^2\le 2$.
Dấu '=' xảy ra khi $x=1$.
$\Rightarrow$ VT $=$ VP $\Leftrightarrow x=1$ (TM)
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