Giải thích các bước giải:
Từ phương trình ta suy ra ;
$\to \sqrt{8x+1}-\sqrt{2x-2}=\sqrt{7x+4}-\sqrt{3x-5}$
$\to (\sqrt{8x+1}-\sqrt{2x-2})^2=(\sqrt{7x+4}-\sqrt{3x-5})^2$
$\to 8x+1+2x-2-2\sqrt{8x+1}.\sqrt{7x+4}=7x+4+3x-5-2\sqrt{7x+4}.\sqrt{3x-5}$
$\to \sqrt{8x+1}.\sqrt{7x+4}=\sqrt{7x+4}.\sqrt{3x-5}$
$\to \sqrt{8x+1}.\sqrt{7x+4}-\sqrt{7x+4}.\sqrt{3x-5}=0$
$\to \sqrt{8x+1}.\sqrt{7x+4}-\sqrt{7x+4}.\sqrt{3x-5}=0$
$\to \sqrt{7x+4}(\sqrt{8x+1}-\sqrt{3x-5})=0$
$\to \sqrt{7x+4}=0$
$\to x=\dfrac{-4}7$
Hoặc $\sqrt{8x+1}-\sqrt{3x-5}=0\to \sqrt{8x+1}=\sqrt{3x-5}\to 8x+1=3x-5\to x=-\dfrac65$