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4 tháng 9 2020

a) \(\sqrt{60}-\sqrt{135}+\frac{1}{3}\sqrt{15}\)

\(=2\sqrt{15}-3\sqrt{15}+\frac{1}{3}\sqrt{15}\)

\(=-\frac{2}{3}\sqrt{15}\)

b) \(\sqrt{28}-\frac{1}{2}\sqrt{343}+2\sqrt{63}\)

\(=2\sqrt{7}-\frac{7}{2}\sqrt{7}+6\sqrt{7}\)

\(=\frac{9}{2}\sqrt{7}\)

c) \(\sqrt{12}-\frac{2}{3}\sqrt{27}+\sqrt{243}\)

\(=2\sqrt{3}-2\sqrt{3}+9\sqrt{3}\)

\(=9\sqrt{3}\)

11) \(\frac{3}{\sqrt{6}-\sqrt{3}}+\frac{4}{\sqrt{7}+\sqrt{3}}\) 12) \(\frac{6}{3\sqrt{2}+2\sqrt{3}}\) 13) \(\left(\sqrt{75}-3\sqrt{2}-\sqrt{12}\right)\left(\sqrt{3}+\sqrt{2}\right)\) 14)\(\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\) 15)\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}\) 16)\(\frac{\sqrt{2}}{2\sqrt{3}+4\sqrt{2}}\) 17)...
Đọc tiếp

11) \(\frac{3}{\sqrt{6}-\sqrt{3}}+\frac{4}{\sqrt{7}+\sqrt{3}}\)

12) \(\frac{6}{3\sqrt{2}+2\sqrt{3}}\)

13) \(\left(\sqrt{75}-3\sqrt{2}-\sqrt{12}\right)\left(\sqrt{3}+\sqrt{2}\right)\)

14)\(\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)

15)\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}\)

16)\(\frac{\sqrt{2}}{2\sqrt{3}+4\sqrt{2}}\)

17) \(\frac{1}{4-3\sqrt{2}}-\frac{1}{4+3\sqrt{2}}\)

18)\(\frac{6}{\sqrt{2}-\sqrt{3}+3}\)

19)\(\frac{\sqrt{3+2\sqrt{2}}+\sqrt{3-2\sqrt{2}}}{\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}}\)

20)\(\sqrt{24}+6\sqrt{\frac{2}{3}}+\frac{10}{\sqrt{6}-1}\)

21)\(2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{58}}\)

22)\(4\sqrt{20}-3\sqrt{125}+5\sqrt{45}-15\sqrt{\frac{1}{5}}\)

23)\(\left(3\sqrt{8}-2\sqrt{12}+\sqrt{20}\right):\left(3\sqrt{18}-2\sqrt{27}+\sqrt{45}\right)\)

24)\(\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)

25)\(\left(\sqrt{7}-\sqrt{5}\right)^2+2\sqrt{35}\)

26)\(\frac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}+\frac{3\sqrt{45}+\sqrt{243}}{\sqrt{5}+\sqrt{3}}\)

27)\(\frac{1}{\sqrt{7-\sqrt{24}}+1}-\frac{1}{\sqrt{7+\sqrt{24}}-1}\)

28)\(\frac{1}{2+\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{2}{3+\sqrt{3}}\)

29)\(\frac{3+\sqrt{5}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{3-\sqrt{5}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)

30)\(\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}\)

31)\(\left(\frac{2}{\sqrt{3}-1}+\frac{3}{\sqrt{3}-2}+\frac{15}{3-\sqrt{3}}\right).\frac{1}{\sqrt{3}+5}\)

32)\(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}-\sqrt{10}\)

3
29 tháng 9 2019

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29 tháng 9 2019

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Bài 1:

a) Ta có: \(\sqrt{243}-\frac{1}{2}\sqrt{12}-2\sqrt{75}+\sqrt{27}\)

\(=\sqrt{3}\cdot9-\frac{1}{2}\cdot\sqrt{3}\cdot2-2\cdot\sqrt{3}\cdot5+\sqrt{3}\cdot3\)

\(=\sqrt{3}\left(9-1-10+3\right)\)

\(=\sqrt{3}\cdot1=\sqrt{3}\)

b) Ta có: \(\frac{2\sqrt{3}-3\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\frac{5}{1+\sqrt{6}}-6\sqrt{\frac{1}{6}}\)

\(=\frac{\left(2\sqrt{3}-3\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}{\left(\sqrt{3}-\sqrt{2}\right)\cdot\left(\sqrt{3}+\sqrt{2}\right)}+\frac{5\cdot\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}-\sqrt{36\cdot\frac{1}{6}}\)

\(=-\sqrt{6}+\frac{5\left(\sqrt{6}-1\right)}{5}-\sqrt{6}\)

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

\(=-\sqrt{6}-1\)

Bài 2: Rút gọn

Ta có: \(\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)

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

\(=\frac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

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

\(=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\frac{3\sqrt{x}}{\sqrt{x}+2}\)

a) Ta có: \(\frac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}-\frac{5-2\sqrt{5}}{2\sqrt{5}-4}\)

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

\(=\sqrt{5}-\frac{\sqrt{5}}{2}\)

\(=\frac{2\sqrt{5}-\sqrt{5}}{2}=\frac{\sqrt{5}}{2}\)

b) Ta có: \(\frac{2\sqrt{8}-\sqrt{12}}{\sqrt{3}-1}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{30}-\sqrt{2}}\)

\(=\frac{\left(2\sqrt{8}-\sqrt{12}\right)\left(\sqrt{3}+1\right)}{3-1}-\frac{\left(\sqrt{5}+\sqrt{27}\right)\left(\sqrt{30}+\sqrt{2}\right)}{30-2}\)

\(=\frac{4\sqrt{6}+4\sqrt{2}-6-2\sqrt{3}}{2}-\frac{5\sqrt{6}+\sqrt{10}+9\sqrt{10}+3\sqrt{6}}{28}\)

\(=\frac{4\sqrt{2}\left(\sqrt{3}+1\right)-2\sqrt{3}\left(\sqrt{3}+1\right)}{2}-\frac{10\sqrt{10}+8\sqrt{6}}{28}\)

\(=\frac{2\cdot\left(\sqrt{3}+1\right)\left(2\sqrt{2}-\sqrt{3}\right)}{2}-\frac{\sqrt{1000}+\sqrt{384}}{28}\)

\(=2\sqrt{6}-3+2\sqrt{2}-\sqrt{3}-\frac{\sqrt{2}\cdot\left(5\sqrt{5}+4\sqrt{3}\right)}{14}\)

\(=\frac{28\sqrt{6}-42+28\sqrt{2}-14\sqrt{3}-10\sqrt{10}-8\sqrt{6}}{14}\)

\(=\frac{20\sqrt{6}-42+28\sqrt{2}-14\sqrt{3}-10\sqrt{10}}{14}\)

c) Ta có: \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)

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

\(=\frac{2\sqrt{3}+2-2\sqrt{3}+2}{2}\)

\(=\frac{4}{2}=2\)