Question

Rút gọn biểu thức sau :

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

a/C/m A = \(\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

b/ Tìm cá giá trị của x để 2P = 2\(\sqrt{x}+5\)

Answer 1

a) ĐKXĐ: \(x\ge0;\ne1\)

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

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

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

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}\) (đpcm)

Vậy \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}}\).

b) Ta có:

\(2P=2\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=2\sqrt{x}+5\)

\(\Rightarrow2\sqrt{x}+2=2x+5\sqrt{x}\)

\(\Leftrightarrow2x+3\sqrt{x}-2=0\)

\(\Leftrightarrow\left(2\sqrt{x}-1\right)\left(\sqrt{x}+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{1}{2}\\\sqrt{x}=-2\left(\text{loại}\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{4}\left(TM\right)\)

Vậy để \(2P=2\sqrt{x}+5\) thì \(x=\dfrac{1}{4}\).

Answer 2

 

a: \(A=\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

b: P là biểu thức nào vậy bạn?