Question

Giải giúp mình bài này với ạ

Answer 1

Lời giải:
ĐKXĐ: $x>0; x\neq 1$

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

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

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

b.

$x=7-4\sqrt{3}=(2-\sqrt{3})^2\Rightarrow \sqrt{x}=2-\sqrt{3}$

Khi đó:

$P=\frac{6-4\sqrt{3}}{2-\sqrt{3}}=-2\sqrt{3}$

c.

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

$\Rightarrow 2(x-1)=3\sqrt{x}$

$\Leftrightarrow 2x-3\sqrt{x}-2=0$

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

$\Rightarrow x=4$ (tm)

Answer 2

Lời giải:
ĐKXĐ: $x>0; x\neq 1$

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

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

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

b.

$x=7-4\sqrt{3}=(2-\sqrt{3})^2\Rightarrow \sqrt{x}=2-\sqrt{3}$

Khi đó:

$P=\frac{6-4\sqrt{3}}{2-\sqrt{3}}=-2\sqrt{3}$

c.

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

$\Rightarrow 2(x-1)=3\sqrt{x}$

$\Leftrightarrow 2x-3\sqrt{x}-2=0$

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

$\Rightarrow x=4$ (tm)