Question

Cho biểu thức:

\(\\ A=\left(\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{2-\sqrt{x}}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\right):\left(2-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)

a) Rút gọn A

b) Tìm x để \(\dfrac{1}{A}\le\dfrac{1}{5}\)

Answer 1

ĐKXĐ : \(x\ge0\) ; \(x\ne4\) và \(x\ne9\)

\(A=\left(\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{2-\sqrt{x}}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\right):\left(2-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)

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

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

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

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

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

Câu b : \(\dfrac{1}{A}< \dfrac{1}{5}\Leftrightarrow\dfrac{x-4}{\sqrt{x}+1}< \dfrac{1}{5}\Leftrightarrow5x-20< \sqrt{x}+1\Leftrightarrow5x-\sqrt{x}-21< 0\)Mysterious Person Tới đây làm sao nữa :(((

Answer 2

\(\text{a) }ĐKXĐ:x\ge0;x\ne4;x\ne9\\ A=\left(\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{2-\sqrt{x}}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\right):\left(2-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\\ =\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right):\left(\dfrac{2\left(\sqrt{x}+1\right)}{\sqrt{x}+1}-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\\ =\dfrac{\sqrt{x}+2+x-9-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}+2-\sqrt{x}}{\sqrt{x}+1}\\ =\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\\ =\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{\sqrt{x}+1}{x-4}\)

\(b\text{) }\dfrac{1}{A}\le\dfrac{1}{5}\\ \Leftrightarrow A\ge5\\ \Leftrightarrow\dfrac{\sqrt{x}+1}{x-4}\ge5\\ \Leftrightarrow\dfrac{\sqrt{x}+1}{x-4}-5\ge0\\ \Leftrightarrow\dfrac{\sqrt{x}+1-5\left(x-4\right)}{x-4}\ge0\\ \Leftrightarrow\dfrac{\sqrt{x}+1-5x+20}{x-4}\ge0\\ \Leftrightarrow\dfrac{\sqrt{x}-5x+21}{x-4}\ge0\\ \Leftrightarrow\dfrac{\left(\sqrt{x}-\dfrac{1+\sqrt{421}}{10}\right)\left(\sqrt{x}-\dfrac{1-\sqrt{421}}{10}\right)}{\sqrt{x}-2}\ge0\)

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