Question

Cho A= [1-\(\dfrac{\sqrt{x}}{1+\sqrt{x}}\)]:[\(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3+\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\)]

a. Tìm điều kiện xác định của A và rút gọn A.

b. Tìm x để A < 0.

c. Tìm x để A=\(\dfrac{1}{2}\)

Answer 1

Ta có A=\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\) với x≥ 9, x ∈ R

Để A > 0 \(\Leftrightarrow\) \(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\) > 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}-2>0\\\sqrt{x}+1< 0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}-2< 0\\\sqrt{x}+1>0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}>2\\\sqrt{x}< -1\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}< 2\\\sqrt{x}>-1\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>4\\x< 1\end{matrix}\right.\\\left\{{}\begin{matrix}x< 4\\x>1\end{matrix}\right.\end{matrix}\right.\)

Kết hợp với ĐKXĐ\(\Rightarrow\) x ∈ ∅

Answer 2

ĐKXĐ: x≥9, x∈R

Ta có:

A= \(\left[\dfrac{1+\sqrt{x}-\sqrt{x}}{1+\sqrt{x}}\right]\):\(\left[\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}+2}{x-2\sqrt{x}-3\sqrt{x}+6}\right]\)

= \(\left[\dfrac{1}{1+\sqrt{x}}\right]\):\(\left[\dfrac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)

=\(\left[\dfrac{1}{1+\sqrt{x}}\right]\):\(\left[\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)

=\(\left[\dfrac{1}{1+\sqrt{x}}\right]\):\(\left[\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)

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

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