Thực hiện phép tính ;
\(A=\frac{\sqrt{2+\sqrt{3}}+\sqrt{4-\sqrt{5}}+\sqrt{10}}{\sqrt{23+2\sqrt{5}}}+\frac{1}{\sqrt{2}+1}\)
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Q
25 tháng 6 2017
a) \(\dfrac{1}{\sqrt{3}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{\sqrt{3}}\sqrt{\dfrac{5}{12}-\dfrac{1}{\sqrt{6}}}\)
\(=\dfrac{\sqrt{3}}{3}+\dfrac{\sqrt{2}}{6}+\dfrac{1}{\sqrt{3}}\sqrt{\dfrac{5}{12}-\dfrac{\sqrt{6}}{6}}\)
\(=\dfrac{\sqrt{3}}{3}+\dfrac{\sqrt{2}}{6}+\dfrac{\sqrt{\dfrac{5}{12}-\dfrac{\sqrt{6}}{6}}}{\sqrt{3}}\)
\(=\dfrac{\sqrt{3}}{3}+\dfrac{\sqrt{2}}{6}+\dfrac{\sqrt{\left(\dfrac{5}{12}-\dfrac{\sqrt{6}}{6}\right)\cdot3}}{3}\)
\(=\dfrac{\sqrt{3}}{3}+\dfrac{\sqrt{2}}{6}+\dfrac{\sqrt{\dfrac{5}{4}-\dfrac{\sqrt{6}}{2}}}{3}\)
\(=\dfrac{\sqrt{3}+\sqrt{\dfrac{5}{4}-\dfrac{\sqrt{6}}{2}}}{3}+\dfrac{\sqrt{2}}{6}\)
b) \(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}=...\)
c) \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}=...\)
d) \(\dfrac{2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\sqrt{3+\sqrt{13+4\sqrt{3}}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\sqrt{3+\sqrt{\left(1+2\sqrt{3}\right)^2}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\sqrt{3+1+2\sqrt{3}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\sqrt{3+2\sqrt{3}+1}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\left(\sqrt{3}+1\right)}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\left(\sqrt{3}+1\right)}\cdot\left(\sqrt{6}+\sqrt{2}\right)}{4}\)
\(=\dfrac{\sqrt{3\left(\sqrt{3}+1\right)}\cdot\left(\sqrt{6}+\sqrt{2}\right)}{2}\)
\(=\dfrac{\sqrt{3-\sqrt{3}-1}\sqrt{\left(\sqrt{6}+\sqrt{2}\right)^2}}{2}\)
\(=\dfrac{\sqrt{\left(3-\sqrt{3}-1\right)\cdot\left(\sqrt{6}+\sqrt{2}\right)^2}}{2}\)
\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\cdot\left(6+2\sqrt{12}+2\right)}}{2}\)
\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\cdot\left(6+4\sqrt{3}+2\right)}}{2}\)
\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\cdot\left(8+4\sqrt{3}\right)}}{2}\)
\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\cdot4\left(2+\sqrt{3}\right)}}{2}\)
\(=\dfrac{\sqrt{\left(4-3\right)\cdot4}}{2}\)
\(=\dfrac{\sqrt{1\cdot4}}{2}\)
\(=\dfrac{2}{2}\)
\(=1\)
18 tháng 8 2016
a/\(\frac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}=\frac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}=2\sqrt{5}+\frac{8}{1-\sqrt{5}}\)
\(=\frac{2\sqrt{5}-10+8}{1-\sqrt{5}}=\frac{-2\left(1-\sqrt{5}\right)}{1-\sqrt{5}}=-2\)
b/Đề sai
c/\(\frac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}=\frac{\sqrt{2}}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\frac{\sqrt{2}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\frac{\sqrt{2}}{3+\sqrt{3}}+\frac{\sqrt{2}}{3-\sqrt{3}}=\sqrt{2}\left(\frac{3+\sqrt{3}+3-\sqrt{3}}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\right)=\frac{6\sqrt{2}}{6}=\sqrt{2}\)
d/ \(\frac{\left(\sqrt{5}+2\right)^2-8\sqrt{5}}{2\sqrt{5}-4}=\frac{9+4\sqrt{5}-8\sqrt{5}}{2\sqrt{5}-4}=\frac{9-4\sqrt{5}}{2\left(\sqrt{5}-2\right)}=\frac{\left(\sqrt{5}-2\right)^2}{2\left(\sqrt{5}-2\right)}=\frac{\sqrt{5}-2}{2}\)
21 tháng 7 2019
a) \(\frac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}\)
= \(\frac{\left(10+2\sqrt{10}\right)\left(1-\sqrt{5}\right)}{\left(\sqrt{5}+\sqrt{2}\right)\left(1-\sqrt{5}\right)}+\frac{8\left(\sqrt{5}+\sqrt{2}\right)}{\left(1-\sqrt{5}\right)\left(\sqrt{5}+\sqrt{2}\right)}\)
= \(\frac{\left(10+2\sqrt{10}\right)\left(1-\sqrt{5}\right)+8\left(\sqrt{5}+\sqrt{2}\right)}{\left(1-\sqrt{5}\right)\left(\sqrt{5}+\sqrt{2}\right)}\)
= \(\frac{10-2\sqrt{5}+2\sqrt{10}-2\sqrt{2}}{\sqrt{5}+\sqrt{2}-5-\sqrt{10}}\)
= \(\frac{2\left(5-\sqrt{5}+\sqrt{10}-\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}-5-\sqrt{10}}\)
= -2
b); c); d) làm tương tự
TV
18 tháng 8 2016
a, = \(\frac{\sqrt{7}-5}{2}-\frac{2\left(3-\sqrt{7}\right)}{4}+\frac{6\left(\sqrt{7}+2\right)}{\left(\sqrt{7}-2\right)\left(\sqrt{7}+2\right)}-\frac{5\left(4-\sqrt{7}\right)}{\left(4-\sqrt{7}\right)\left(4+\sqrt{7}\right)}\)
13 tháng 7 2016
a) Kết quả rút gọn xấu (+dài) nữa. (có thể đề sai)
b)
\(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}\)
\(=\left[\frac{-\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\frac{-\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right].\left(\sqrt{7}-\sqrt{5}\right)\)
\(=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)=-\left(7-5\right)=-2\)
c) \(\frac{\sqrt{5-2\sqrt{6}}+\sqrt{8-2\sqrt{15}}}{\sqrt{7+2\sqrt{10}}}=\frac{\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}}{\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}}\)
\(=\frac{\sqrt{3}-\sqrt{2}+\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{2}}=\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}=\frac{\left(\sqrt{5}-\sqrt{2}\right)^2}{3}\)
14 tháng 7 2016
a) \(\left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right).\frac{1}{\sqrt{6}}=\left[\frac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-2\sqrt{6}\right].\frac{1}{\sqrt{6}}\)
\(=\left(\frac{\sqrt{6}}{2}-2\sqrt{6}\right).\frac{1}{\sqrt{6}}=\frac{1}{2}-2=-\frac{3}{2}\)
NN
26 tháng 10 2020
a) \(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{1+\sqrt{2}}-2+\sqrt{3}\)
\(=\frac{\sqrt{3}.\left(\sqrt{3}+2\right)}{\sqrt{3}}+\frac{\sqrt{2}.\left(\sqrt{2}+1\right)}{1+\sqrt{2}}-2+\sqrt{3}\)
\(=\sqrt{3}+2+\sqrt{2}-2+\sqrt{3}\)
\(=2\sqrt{3}+\sqrt{2}\)
b) \(\frac{-3}{2}.\sqrt{9-4\sqrt{5}}+\sqrt{\left(-4\right)^2.\left(1+\sqrt{5}\right)^2}\)
\(=\frac{-3}{2}.\sqrt{5-4\sqrt{5}+4}+\sqrt{4^2.\left(1+\sqrt{5}\right)^2}\)
\(=\frac{-3}{2}.\sqrt{\left(\sqrt{5}-2\right)^2}+\sqrt{4^2}.\sqrt{\left(1+\sqrt{5}\right)^2}\)
\(=\frac{-3}{2}.\left|\sqrt{5}-2\right|+4.\left|1+\sqrt{5}\right|\)
\(=\frac{-3}{2}.\left(\sqrt{5}-2\right)+4\left(1+\sqrt{5}\right)\)
\(=\frac{-3\sqrt{5}}{2}+3+4+4\sqrt{5}\)
\(=\frac{-3\sqrt{5}}{2}+4\sqrt{5}+7\)
\(=\frac{-3\sqrt{5}}{2}+\frac{8\sqrt{5}}{2}+\frac{14}{2}\)
\(=\frac{-3\sqrt{5}+8\sqrt{5}+14}{2}=\frac{14+5\sqrt{5}}{2}\)