Question

Cho biểu thức: P= \(\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\) 

Rút gọn và tìm giá trị của x để P=\(\dfrac{3}{2}\)

Answer 1

Lời giải:
ĐK: $x>0; x\neq 1$

\(P=\left[\frac{x-2}{\sqrt{x}(\sqrt{x}+2)}+\frac{\sqrt{x}}{\sqrt{x}(\sqrt{x}+2)}\right].\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(=\frac{x+\sqrt{x}-2}{\sqrt{x}(\sqrt{x}+2)}.\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{(\sqrt{x}-1)(\sqrt{x}+2)}{\sqrt{x}(\sqrt{x}+2)}.\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{\sqrt{x}+1}{\sqrt{x}}\)

$P=\frac{3}{2}\Leftrightarrow \frac{\sqrt{x}+1}{\sqrt{x}}=\frac{3}{2}$

$\Leftrightarrow \frac{1}{\sqrt{x}}=\frac{1}{2}$

$\Leftrightarrow x=4$ (thỏa mãn)

Answer 2

ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

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

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

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

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

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

Để \(P=\dfrac{3}{2}\) thì \(2\left(\sqrt{x}+1\right)=3\sqrt{x}\)

\(\Leftrightarrow3\sqrt{x}-2\sqrt{x}=2\)

\(\Leftrightarrow\sqrt{x}=2\)

hay x=4(thỏa ĐK)