Bài 1: Căn bậc hai

Mai Anh

Rút gọn biểu thức :

a) \(\sqrt{\dfrac{2-\sqrt{5}}{\sqrt{5}-3}}:\sqrt{\left(\sqrt{5}-3\right)\left(2-\sqrt{5}\right)}\)

b) \(\dfrac{2+3\sqrt{5}}{\sqrt{5}-2}-\dfrac{\sqrt{5}+1}{\sqrt{5}+2}\)

c) \(\dfrac{1+\sqrt{2}}{\sqrt{4-2\sqrt{3}}}:\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\)

qwerty
13 tháng 6 2017 lúc 10:20

a) \(\sqrt{\dfrac{2-\sqrt{5}}{\sqrt{5}-3}}:\sqrt{\left(\sqrt{5}-3\right)\left(2-\sqrt{5}\right)}\)

\(=\sqrt{\dfrac{2-\sqrt{5}}{\sqrt{5}-3}}:\left(\left(\sqrt{5}-3\right)\cdot\left(2-\sqrt{5}\right)\right)\)

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

\(=\sqrt{\dfrac{2-\sqrt{5}}{\sqrt{5}-3}:\left(5\sqrt{5}-11\right)}\)

\(=\sqrt{\dfrac{2-\sqrt{5}}{\sqrt{5}-3}\cdot\dfrac{1}{5\sqrt{5}-11}}\)

\(=\sqrt{\dfrac{2-\sqrt{5}}{\left(\sqrt{5}-3\right)\cdot\left(5\sqrt{5}-1\right)}}\)

\(=\sqrt{\dfrac{\left(2-\sqrt{5}\right)\cdot\left(\sqrt{5}+3\right)}{-4\left(5\sqrt{5}-1\right)}}\)

\(=\sqrt{\dfrac{2\sqrt{5}+6-5-3\sqrt{5}}{-4\left(5\sqrt{5}-11\right)}}\)

\(=\sqrt{\dfrac{-\sqrt{5}+1}{-4\left(5\sqrt{5}-11\right)}}\)

\(=\sqrt{-\dfrac{\left(-\sqrt{5}+1\right)\cdot\left(5\sqrt{5}+11\right)}{16}}\)

\(=\sqrt{-\dfrac{-25-11\sqrt{5}+5\sqrt{5}+11}{16}}\)

\(=\sqrt{-\dfrac{-14-6\sqrt{5}}{16}}\)

\(=\sqrt{-\dfrac{2\left(-7-3\sqrt{5}\right)}{16}}\)

\(=\sqrt{-\dfrac{-7-3\sqrt{5}}{8}}\)

\(=\dfrac{\sqrt{-\left(-7-3\sqrt{5}\right)}}{\sqrt{8}}\)

\(=\dfrac{\sqrt{7+3\sqrt{5}}}{2\sqrt{2}}\)

\(=\dfrac{\sqrt{\left(7+3\sqrt{5}\right)\cdot2}}{4}\)

\(=\dfrac{\sqrt{14+6\sqrt{5}}}{4}\)

\(=\dfrac{\sqrt{\left(3+\sqrt{5}\right)^2}}{4}\)

\(=\dfrac{3+\sqrt{5}}{4}\)

b) \(\dfrac{2+3\sqrt{5}}{\sqrt{5}-2}-\dfrac{\sqrt{5}+1}{\sqrt{5}+2}\)

\(=\left(2+3\sqrt{5}\right)\cdot\left(\sqrt{5}+2\right)-\left(\sqrt{5}+1\right)\cdot\left(\sqrt{5}-2\right)\)

\(=2\sqrt{5}+4+15+6\sqrt{5}-\left(5-2\sqrt{5}+\sqrt{5}-2\right)\)

\(=2\sqrt{5}+4+15+6\sqrt{5}-\left(3-\sqrt{5}\right)\)

\(=2\sqrt{5}+4+15+6\sqrt{5}-3+\sqrt{5}\)

\(=9\sqrt{5}+16\)

c) \(\dfrac{1+\sqrt{2}}{\sqrt{4-2\sqrt{3}}}:\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\)

\(=\dfrac{1+\sqrt{2}}{\sqrt{\left(1-\sqrt{3}\right)^2}}\cdot\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\)

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

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

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

\(=\dfrac{2-1}{2}\)

\(=\dfrac{1}{2}\)

Bình luận (3)
Mysterious Person
13 tháng 6 2017 lúc 10:42

a) \(\sqrt{\dfrac{2-\sqrt{5}}{\sqrt{5}-3}}:\sqrt{\left(\sqrt{5}-3\right)\left(2-\sqrt{5}\right)}\)= \(\dfrac{\sqrt{2-\sqrt{5}}}{\sqrt{\sqrt{5}-3}}.\dfrac{1}{\sqrt{\sqrt{5}-3}\sqrt{2-\sqrt{5}}}\)

= \(\dfrac{1}{\sqrt{\sqrt{5}-3}}.\dfrac{1}{\sqrt{\sqrt{5}-3}}\) = \(\dfrac{1}{\sqrt{\sqrt{5}-3}^2}\) = \(\dfrac{1}{3-\sqrt{5}}\)

b) \(\dfrac{2+3\sqrt{5}}{\sqrt{5}-2}-\dfrac{\sqrt{5}+1}{\sqrt{5}+2}\) = \(\dfrac{\left(2+3\sqrt{5}\right)\left(\sqrt{5}+2\right)-\left(\sqrt{5}+1\right)\left(\sqrt{5}-2\right)}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}\)

= \(\dfrac{2\sqrt{5}+4+15+6\sqrt{5}-\left(5-2\sqrt{5}+\sqrt{5}-2\right)}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}\)

= \(\dfrac{8\sqrt{5}+19-5+2\sqrt{5}-\sqrt{5}+2}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}\) = \(\dfrac{9\sqrt{5}+16}{5-4}\) = \(9\sqrt{5}+16\)

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

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

= \(\dfrac{1}{2}\)

Bình luận (0)

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