Question

Cho biểu thức A=\(\dfrac{1}{2\sqrt{X}-2}-\dfrac{1}{2\sqrt{X}+2}+\dfrac{\sqrt{X}}{1-X}\)

a) Rút gọn A

b) Tính giá trị của A khi x=\(\dfrac{4}{9}\)

c) Tìm x để A=\(\left(\dfrac{-1}{2};\dfrac{-1}{4}\right)\)

Answer 1

a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

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

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

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

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

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

b) Khi \(x=\dfrac{4}{9}\) (thảo mãn ĐKXĐ) thì giá trị của A là:

\(A=-\dfrac{1}{\sqrt{x}+1}=-\dfrac{1}{\sqrt{\dfrac{4}{9}}+1}=-\dfrac{3}{5}\)

Vậy .....

c)

+) Khi \(A=-\dfrac{1}{2}\) thì ta có:

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

\(\Leftrightarrow x=1\) (Loại do không thỏa mãn ĐKXĐ)

+) Khi \(A=\dfrac{-1}{4}\) thì ta có:

\(A=-\dfrac{1}{\sqrt{x}+1}=-\dfrac{1}{4}\)

\(\Leftrightarrow x=9\) (thỏa mãn)

Vậy để A = \(-\dfrac{1}{4}\) thì x = 9

Answer 2

a/ ĐKXĐ: \(x\ge0,x\ne1\)

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

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

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

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

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

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

b/

Thay \(x=\dfrac{4}{9}\) vào A ta được:

\(A=\dfrac{-1}{\sqrt{\dfrac{4}{9}}+1}=\dfrac{-1}{\dfrac{2}{3}+1}=\dfrac{-3}{5}\)

Vậy khi \(x=\dfrac{4}{9}\) thì \(A=\dfrac{-3}{5}\)

c/ Với \(x\ge0,x\ne1\)

* Để \(A=\dfrac{-1}{2}\Leftrightarrow\dfrac{-1}{\sqrt{x}+1}=\dfrac{-1}{2}\)

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

\(\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\) ( ktmđk)-Loại

Vậy không có giá trị nào của x thỏa mãn \(A=\dfrac{-1}{2}\)

* Để \(A=\dfrac{-1}{4}\Leftrightarrow\dfrac{-1}{\sqrt{x}+1}=\dfrac{-1}{4}\)

\(\Leftrightarrow-4=-\sqrt{x}-1\)

\(\Leftrightarrow\sqrt{x}=3\Leftrightarrow x=9\) (tmđk)

Vậy để \(A=\dfrac{-1}{4}\) thì \(x=9\)