Đáp án: $x=12$
Giải thích các bước giải:
Ta có:
$\dfrac{(2x)^2}{(1+\sqrt{2x+1})^2}=2x-8$
$\to \dfrac{(2x)^2(\sqrt{2x+1}-1)^2}{((1+\sqrt{2x+1})(\sqrt{2x+1}-1))^2}=2x-8$
$\to \dfrac{(2x)^2(\sqrt{2x+1}-1)^2}{(2x+1-1)^2}=2x-8$
$\to \dfrac{(2x)^2(\sqrt{2x+1}-1)^2}{(2x)^2}=2x-8$
$\to (\sqrt{2x+1}-1)^2=2x-8$
$\to 2x+1-2\sqrt{2x+1}+1=2x-8$
$\to -2\sqrt{2x+1}=-10$
$\to \sqrt{2x+1}=5$
$\to 2x+1=25$
$\to x=12$