Chương I - Căn bậc hai. Căn bậc ba

HT

Rút gọn

A=\(\dfrac{\sqrt{2}}{\sqrt{3}+\sqrt{5}}+\dfrac{2\sqrt{2}}{\sqrt{10}-\sqrt{2}}\)

B=\(\left(\dfrac{\sqrt{a}}{1-\sqrt{a}}+\dfrac{\sqrt{a}}{a-1}\right):\dfrac{\sqrt{a}}{a-1}\)

H24
7 tháng 7 2017 lúc 16:02

1) \(A=\dfrac{\sqrt{2}}{\sqrt{3}+\sqrt{5}}+\dfrac{2\sqrt{2}}{\sqrt{10}-\sqrt{2}}\)

\(=\dfrac{\sqrt{2}\left(\sqrt{3}-\sqrt{5}\right)}{-2}+\dfrac{2\sqrt{2}\left(\sqrt{10}+\sqrt{2}\right)}{8}\)

\(=\dfrac{\sqrt{6}-\sqrt{10}}{-2}+\dfrac{\sqrt{2}\left(\sqrt{10}+\sqrt{2}\right)}{4}\)

\(=-\dfrac{\sqrt{6}-\sqrt{10}}{2}+\dfrac{\sqrt{20}+2}{4}\)

\(=-\dfrac{\sqrt{6}-\sqrt{10}}{2}+\dfrac{2\sqrt{5}+2}{4}\)

\(=-\dfrac{\sqrt{6}-\sqrt{10}}{2}+\dfrac{2\left(\sqrt{5}+1\right)}{4}\)

\(=-\dfrac{\sqrt{6}-\sqrt{10}}{2}+\dfrac{\sqrt{5}+1}{2}\)

\(=\dfrac{-\left(\sqrt{6}-\sqrt{10}\right)+\sqrt{5}+1}{2}\)

\(=\dfrac{-\sqrt{6}+\sqrt{10}+\sqrt{5}+1}{2}\)

2) \(\left(\dfrac{\sqrt{a}}{1-\sqrt{a}}+\dfrac{\sqrt{a}}{a-1}\right):\dfrac{\sqrt{a}}{a-1}\)

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

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

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

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

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

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

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

\(=-a\cdot\dfrac{1}{\sqrt{a}}\)

\(=-\dfrac{a}{\sqrt{a}}\)

\(=-\dfrac{a\sqrt{a}}{a}\)

\(=-\sqrt{a}\)

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