Đáp án: $x=169$
Giải thích các bước giải:
ĐKXĐ: $x\ge 0, x\ne 1$
Ta có:
$\dfrac{\sqrt{x}-3}{\sqrt{x}+2}=2-\dfrac{\sqrt{x}+3}{\sqrt{x}-1}$
$\to \dfrac{\sqrt{x}-3}{\sqrt{x}+2}-1=1-\dfrac{\sqrt{x}+3}{\sqrt{x}-1}$
$\to \dfrac{\sqrt{x}-3-(\sqrt{x}+2)}{\sqrt{x}+2}=\dfrac{\sqrt{x}-1-(\sqrt{x}+3)}{\sqrt{x}-1}$
$\to \dfrac{-5}{\sqrt{x}+2}=\dfrac{-4}{\sqrt{x}-1}$
$\to \dfrac{5}{\sqrt{x}+2}=\dfrac{4}{\sqrt{x}-1}$
$\to 5(\sqrt{x}-1)=4(\sqrt{x}+2)$
$\to 5\sqrt{x}-5=4\sqrt{x}+8$
$\to \sqrt{x}=13$
$\to x=169$
