Question

[manh]

rút gọn biểu thức 

\(\dfrac{x+12}{x-4}+\dfrac{1}{\sqrt{x}+2}-\dfrac{4}{\sqrt{x}-2}\) x≥0

Answer 1

[Nguyễn Lê Phước Thịnh]

ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >4\end{matrix}\right.\)

\(\dfrac{x+12}{x-4}+\dfrac{1}{\sqrt{x}+2}-\dfrac{4}{\sqrt{x}-2}\)

\(=\dfrac{x+12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}+2}-\dfrac{4}{\sqrt{x}-2}\)

\(=\dfrac{x+12+\sqrt{x}-2-4\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{x+\sqrt{x}+10-4\sqrt{x}-8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{x-3\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)