Question

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

a. Rút gọn A

b. Tìm x để A = 2

Answer 1

a) \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{x}{x-1}\right):\left(\dfrac{2x}{x-1}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\left(x\ge0,x\ne1\right)\)

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

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

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

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

b) \(A=2\Rightarrow\dfrac{-1}{\sqrt{x}-1}=2\Rightarrow-1=2\sqrt{x}-2\Rightarrow2\sqrt{x}=1\Rightarrow\sqrt{x}=\dfrac{1}{2}\)

\(\Rightarrow x=\dfrac{1}{4}\)

Answer 2

Lời giải:

ĐK: $x\geq 0; x\neq 1$

a. 

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

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

b.

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

$\Leftrightarrow \sqrt{x}=\frac{1}{2}\Leftrightarrow x=\frac{1}{4}$ (tm)

 

Answer 3

A=(\(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{x}{x-1}\) )\(\div\left(\dfrac{2x}{x-1}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\) (ĐK  \(x\ge0,x\ne1\) )

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

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

  =\(\dfrac{-1}{\sqrt{x}-1}\)

câu b

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

Answer 4

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

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

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

b) Để A=2 thì \(\sqrt{x}-1=\dfrac{-1}{2}\)

hay \(x=\dfrac{1}{4}\)