Question

Cho biểu thức A = \(\left(\dfrac{2\sqrt{x}}{x-9}+\dfrac{1}{\sqrt{x}-3}\right)\): \(\dfrac{3}{\sqrt{x}-3}\) với x \(\ge0;\) \(x\ne9\)

1) Rút gọn biểu thức A

2) Tìm x để A = \(\dfrac{5}{6}\)

3) Tìm giá trị nhỏ nhất của biểu thức A

Answer 1

1) ĐKXĐ: \(\left\{{}\begin{matrix}\sqrt{x}\ge0\\x-9\ne0\\\sqrt{x}-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)\(A=\left(\dfrac{2\sqrt{x}}{x-9}+\dfrac{1}{\sqrt{x}-3}\right):\dfrac{3}{\sqrt{x}-3}=\dfrac{2\sqrt{x}+\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{3}=\dfrac{3\sqrt{x}+3}{3\left(\sqrt{x}+3\right)}=\dfrac{3\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+3\right)}\)2) Để A=\(\dfrac{5}{6}\) thì \(\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+3\right)}=\dfrac{5}{6}\Leftrightarrow\left(\sqrt{x}+1\right)6=\left(\sqrt{x}+3\right)5\Leftrightarrow6\sqrt{x}+6=5\sqrt{x}+15\Leftrightarrow\sqrt{x}=9\Leftrightarrow x=81\)

Answer 2

1. Ta có:

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

\(=\dfrac{2\sqrt{x}.\left(\sqrt{x}-3\right)}{3\left(x-9\right)}+\dfrac{1}{3}\)

\(=\dfrac{2x-6\sqrt{x}}{3\left(x-9\right)}+\dfrac{x-9}{3\left(x-9\right)}\)

\(=\dfrac{3x-6\sqrt{x}-9}{3x-27}\)

\(=\dfrac{x-2\sqrt{x}-3}{x-9}\)