Giải thích các bước giải:
Ta có :
$y=\sqrt{x-1}+\sqrt{4-2x}$
$\rightarrow y=\dfrac{1}{\sqrt{2}}.\sqrt{2x-2}+\sqrt{4-2x}$
$\rightarrow y^2=(\dfrac{1}{\sqrt{2}}.\sqrt{2x-2}+1.\sqrt{4-2x})^2$
$\rightarrow y^2=(\dfrac{1}{2}+1)(2x-2+4-2x)=3$
$\rightarrow y\le \sqrt{3}$
Dấu = xảy ra khi $\dfrac{\sqrt{2x-2}}{\dfrac{1}{\sqrt{2}}}=\dfrac{\sqrt{4-2x}}{1}\rightarrow x=\dfrac{4}{3}$