Question

Cho phương trình :

\(P=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\right)+\left(\dfrac{2\sqrt{x}}{\sqrt{x}+2}\right)+\left(\dfrac{2+5\sqrt{x}}{4-x}\right)\)

a, Tìm \(ĐKXĐ\) của pt

b, Rút gọn pt

c, Tìm x để P = 2

Answer 1

a. \(ĐKXĐ\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)

b. \(P=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}\)

\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+2\sqrt{x}\left(\sqrt{x}-2\right)-2-5\sqrt{x}}{x-4}\)

\(=\dfrac{3x-6\sqrt{x}}{x-4}=\dfrac{3\sqrt{x}\left(\sqrt{x-2}\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)

c. \(P=2\Leftrightarrow\dfrac{3\sqrt{x}}{\sqrt{x}+2}=2\Leftrightarrow3\sqrt{x}=2\sqrt{x}+4\Leftrightarrow\sqrt{x}=4\Leftrightarrow x=16\)

Vậy x = 16