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31 tháng 5 2017
a ) \(\dfrac{2}{\sqrt{3}-1}\) - \(\dfrac{2}{\sqrt{3}+1}\) = \(\dfrac{2\left(\sqrt{3}+1\right)-2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
= \(\dfrac{2\sqrt{3}+2-2\sqrt{3}+2}{3-1}\) = \(\dfrac{4}{2}\) = 2
b) \(\dfrac{5}{12\left(2\sqrt{5}+3\sqrt{2}\right)}\) - \(\dfrac{5}{12\left(2\sqrt{5}-3\sqrt{2}\right)}\)
= \(\dfrac{5\left(2\sqrt{5}-3\sqrt{2}\right)-5\left(2\sqrt{5}+3\sqrt{2}\right)}{12\left(2\sqrt{5}+3\sqrt{2}\right)\left(2\sqrt{5}-3\sqrt{2}\right)}\)
= \(\dfrac{10\sqrt{5}-15\sqrt{2}-10\sqrt{5}-15\sqrt{2}}{12\left(20-18\right)}\)
= \(\dfrac{-30\sqrt{2}}{24}\) = \(\dfrac{-15\sqrt{2}}{12}\) = \(\dfrac{-5\sqrt{2}}{4}\)
c) \(\dfrac{5+\sqrt{5}}{5-\sqrt{5}}\) +\(\dfrac{5-\sqrt{5}}{5+\sqrt{5}}\) = \(\dfrac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\)
= \(\dfrac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{25-5}\) = \(\dfrac{60}{20}\) = 3
31 tháng 5 2017
d) \(\dfrac{\sqrt{3}}{\sqrt{\sqrt{3+1}}-1}\) - \(\dfrac{\sqrt{3}}{\sqrt{\sqrt{3+1}}+1}\)
= \(\dfrac{\sqrt{3}}{\sqrt{2}-1}\) - \(\dfrac{\sqrt{3}}{\sqrt{2}+1}\) = \(\dfrac{\sqrt{3}\left(\sqrt{2}+1\right)-\sqrt{3}\left(\sqrt{2}-1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\)
= \(\dfrac{\sqrt{6}+\sqrt{3}-\sqrt{6}+\sqrt{3}}{2-1}\) = \(2\sqrt{3}\)
15 tháng 7 2017
a) \(\left(2-\sqrt{2}\right)\left(-5\sqrt{2}\right)-\left(3\sqrt{2}-5\right)^2\)
\(=-10\sqrt{2}+5.2-\left(18-30\sqrt{2}+25\right)\)
\(=-10\sqrt{2}+10-18+30\sqrt{2}-25\)
\(=20\sqrt{2}-33\)
b) câu b đề sai
24 tháng 4 2017
a. \(\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{2}.\sqrt{2}}=\dfrac{\sqrt{10}-\sqrt{6}}{2}\)
b. \(\dfrac{26}{5-2\sqrt{3}}=\dfrac{26\left(5+2\sqrt{3}\right)}{\left(5+2\sqrt{3}\right)\left(5-2\sqrt{3}\right)}=\dfrac{26\left(5+2\sqrt{3}\right)}{13}=2\left(5+2\sqrt{3}\right)=10+4\sqrt{3}\)
c. \(\dfrac{2\sqrt{10}-5}{4-\sqrt{10}}=\dfrac{\left(2\sqrt{10}-5\right)\left(4+\sqrt{10}\right)}{\left(4-\sqrt{10}\right)\left(4+\sqrt{10}\right)}=\dfrac{3\sqrt{10}}{6}=\dfrac{\sqrt{10}}{2}\)
d. \(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}=\dfrac{\left(9-2\sqrt{3}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}{\left(3\sqrt{6}-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}=\dfrac{23\sqrt{6}}{46}=\dfrac{\sqrt{6}}{2}\)
KV
0
DH
2 tháng 5 2017
a) \(\dfrac{x^2-5}{x+\sqrt{5}}\)(với x khác -\(\sqrt{5}\)) =\(\dfrac{\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)}{x+\sqrt{5}}\) = x-\(\sqrt{5}\) vậy \(\dfrac{x^2-5}{x+\sqrt{5}}\) = x-\(\sqrt{5}\) với x khác -\(\sqrt{5}\) b) \(\dfrac{x^2+2\sqrt{2}x+2}{x^2-2}\) ( với x khác +-\(\sqrt{2}\) ) = \(\dfrac{\left(x+\sqrt{2}\right)^2}{\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)}\) =\(\dfrac{x+\sqrt{2}}{x-\sqrt{2}}\) vậy \(\dfrac{x^2+2\sqrt{2}x+2}{x^2-2}\) =\(\dfrac{x+\sqrt{2}}{x-\sqrt{2}}\) với x khác +-\(\sqrt{2}\)
15 tháng 7 2017
a) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}=\sqrt{16}-6+\sqrt{20}-\sqrt{5}=4-6+2\sqrt{5}-\sqrt{5}=\sqrt{5}-2\)
b) \(0,2\sqrt{\left(-10\right)^3.3}+2\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}=0,2\left|-10\right|\sqrt{3}+2\left|\sqrt{3}-\sqrt{5}\right|=0,2.10.\sqrt{3}+2\left(\sqrt{5}-\sqrt{3}\right)=2\sqrt{3}+2\sqrt{5}-2\sqrt{3}=2\sqrt{5}\)
c) \(\left(\dfrac{1}{2}\sqrt{\dfrac{1}{2}}-\dfrac{3}{2}\sqrt{2}+\dfrac{4}{5}\sqrt{200}\right):\dfrac{1}{8}=\left(\dfrac{1}{2}\sqrt{\dfrac{2}{4}}-\dfrac{3}{2}\sqrt{2}+8\sqrt{2}\right):\dfrac{1}{8}=\left(\dfrac{1}{4}\sqrt{2}-\dfrac{2}{3}\sqrt{2}+8\sqrt{2}\right):\dfrac{1}{8}=\dfrac{27}{4}\sqrt{2}.8=54\sqrt{2}\)
d) \(2\sqrt{\left(\sqrt{2}-3\right)^2}+\sqrt{2.\left(-3\right)^2}-5\sqrt{\left(-1\right)^4}=2\left(3-\sqrt{2}\right)+3\sqrt{2}-5=6-2\sqrt{2}+3\sqrt{2}-5=1+\sqrt{2}\)
KL
27 tháng 6 2017
1. \(=\sqrt{\left(\sqrt{\frac{7}{2}}+\sqrt{\frac{3}{2}}\right)^2}+\sqrt{\left(\sqrt{\frac{7}{2}}-\sqrt{\frac{3}{2}}\right)^2}-2\sqrt{4\sqrt{7}}=\frac{7}{2}+\frac{3}{2}+\frac{7}{2}-\frac{3}{2}-2\sqrt{4\sqrt{7}}\)
\(=7-2\sqrt{4\sqrt{7}}\)
29 tháng 5 2018
cho hỏi tại sao có số \(\frac{7}{2};\frac{3}{2}\)zậy chỉ với
\(A=\dfrac{5\sqrt{5}+5\sqrt{11}}{5}=\sqrt{5}+\sqrt{11}\)
`A=\sqrt{5}:5/[5+\sqrt{55}]`
`A=\sqrt{5}. [5+\sqrt{55}]/5`
`A=[5\sqrt{5}+5\sqrt{11}]/5`
`A=[5(\sqrt{5}+\sqrt{11})]/5=\sqrt{5}+\sqrt{11}`