Question

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1. Rút gọn M

M= (\(\dfrac{x+3\sqrt{x}}{x-25}+\dfrac{1}{\sqrt{x}+5}\) ) :\(\dfrac{\sqrt{x}+2}{\sqrt{x}-5}\)

2. Tính giá trị của M khi x = \(\dfrac{1}{\sqrt{3}-1}-\dfrac{1}{\sqrt{3}+1}\)

Answer 1

1) ĐKXĐ: \(x\ge0;x\ne25\)

\(M=\left(\dfrac{x+3\sqrt{x}}{x-25}+\dfrac{1}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+2}{\sqrt{x}-5}\)

\(=\dfrac{x+3\sqrt{x}+\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)

\(=\dfrac{x+4\sqrt{x}-5}{\sqrt{x}+5}.\dfrac{1}{\sqrt{x}+2}\)

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

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

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

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

\(=1\) (Thỏa mãn ĐKXĐ của M)

Thay \(x=1\) vào M ta có:

\(M=\dfrac{\sqrt{1}-1}{\sqrt{1}+2}=0\)