Giải thích các bước giải:
đkxđ : $4\le x\le 7$
Ta có :
$\sqrt{x+3}-\sqrt{7-x}>\sqrt{2x-8}$
$\to \sqrt{x+3}>\sqrt{2x-8}+\sqrt{7-x}$
$\to x+3>(\sqrt{2x-8}+\sqrt{7-x})^2$
$\to x+3>2x-8+2.\sqrt{2x-8}.\sqrt{7-x}+7-x$
$\to x+3>x-1+2\sqrt{-2x^2+22x-56}$
$\to 2>2\sqrt{-2x^2+22x-56}$
$\to \sqrt{-2x^2+22x-56}<1$
$\to -2x^2+22x-56<1$
$\to 2x^2-22x+57>0$
$\to 2(x-\dfrac{11}{2})^2>\dfrac{7}{2}$
$\to (x-\dfrac{11}{2})^2>\dfrac{7}{4}$
$\to x<\dfrac{-\sqrt{7}+11}{2}$ hoặc $x>\dfrac{\sqrt{7}+11}{2}$
$\to \dfrac{\sqrt{7}+11}{2}<x\le 7$
