\(\sqrt{85}+\sqrt{5\frac{1}{3}}+2\sqrt{75}-5\sqrt{1\frac{1}{3}}\)
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27 tháng 8 2019
a)\(\sqrt{75}-\sqrt{5\frac{1}{3}}+\frac{9}{2}\sqrt{2\frac{2}{3}}+2\sqrt{27}=5\sqrt{3}-\frac{\sqrt{15}}{3}+3\sqrt{3}+6\sqrt{3}=14\sqrt{3}-\frac{\sqrt{15}}{3}\)
b) \(\sqrt{48}+\sqrt{5\frac{1}{3}}+2\sqrt{75}-5\sqrt{1\frac{1}{3}}=4\sqrt{3}+\frac{\sqrt{15}}{3}+10\sqrt{3}-\frac{5\sqrt{3}}{3}=\frac{12\sqrt{3}+30\sqrt{3}-5\sqrt{3}}{3}+\frac{\sqrt{15}}{3}=\frac{37\sqrt{3}+\sqrt{15}}{3}\)
c) \(\left(\sqrt{15}+2\sqrt{3}\right)^2+12\sqrt{5}=\left[\left(\sqrt{15}\right)^2+4\sqrt{45}+\left(2\sqrt{3}\right)^2\right]+12\sqrt{5}=15+12\sqrt{5}+12+12\sqrt{5}=27+24\sqrt{5}\)
d) \(\left(\sqrt{6}+2\right)\left(\sqrt{3}-\sqrt{2}\right)=\sqrt{18}-\sqrt{12}+\sqrt{6}-2\sqrt{2}=3\sqrt{2}-2\sqrt{3}+\sqrt{6}-2\sqrt{2}=\sqrt{2}-2\sqrt{3}+\sqrt{6}\)
e) \(\left(\sqrt{3}+1\right)^2-2\sqrt{3}+4=\left(\sqrt{3}\right)^2+2\sqrt{3}+1-2\sqrt{3}+4=3+2\sqrt{3}+1-2\sqrt{3}+4=8\)
f) \(\frac{1}{7+4\sqrt{3}}+\frac{1}{7-4\sqrt{3}}=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=\frac{14}{1}=14\)
g) \(\left(\frac{1}{\sqrt{5}-\sqrt{2}}-\frac{1}{\sqrt{5}+\sqrt{2}}+1\right)\frac{1}{\left(\sqrt{2}+1\right)^2}=\left(\frac{\sqrt{5}+2-\sqrt{5}+2+5-2}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}\right)\frac{1}{3+2\sqrt{2}}=\frac{7}{3}.\frac{1}{3+2\sqrt{2}}=\frac{7}{9+6\sqrt{2}}\)
23 tháng 8 2020
1) Ta có: \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+2}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}{\left(\sqrt{2}+\sqrt{3}+2\right)}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+2\right)\left(1+\sqrt{2}\right)}{\left(\sqrt{2}+\sqrt{3}+2\right)}\)
\(=1+\sqrt{2}\)
2) Ta có: \(2\sqrt{27}-6\sqrt{\frac{4}{3}}+\frac{3}{5}\sqrt{75}\)
\(=\sqrt{108}-\sqrt{36\cdot\frac{4}{3}}+\sqrt{75\cdot\frac{9}{25}}\)
\(=\sqrt{108}-\sqrt{48}+\sqrt{27}\)
\(=\sqrt{3}\left(6-4+3\right)\)
\(=5\sqrt{3}\)
3) Sửa đề: \(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{192}\)
Ta có: \(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{192}\)
\(=\sqrt{2}\cdot\sqrt{4}\cdot\sqrt{3}-10\sqrt{4}\cdot\sqrt{3}+16\cdot\sqrt{4}\cdot\sqrt{3}\)
\(=\sqrt{2}\cdot\sqrt{12}-10\sqrt{12}+16\sqrt{12}\)
\(=\sqrt{12}\left(\sqrt{2}-10+16\right)\)
\(=2\sqrt{3}\left(\sqrt{2}-6\right)\)
\(=2\sqrt{6}-12\sqrt{3}\)
4) Ta có: \(\frac{1}{2+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}\)
\(=\frac{2-\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+\frac{\sqrt{12}}{6}-\frac{2\left(3-\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\)
\(=\frac{6\left(2-\sqrt{3}\right)+2\sqrt{3}-6+2\sqrt{3}}{6}\)
\(=\frac{12-6\sqrt{3}+2\sqrt{3}-6+2\sqrt{3}}{6}\)
\(=\frac{6-2\sqrt{3}}{6}\)
\(=\frac{2\sqrt{3}\left(\sqrt{3}-1\right)}{2\sqrt{3}\cdot\sqrt{3}}\)
\(=\frac{\sqrt{3}-1}{\sqrt{3}}\)
5) Ta có: \(\left(\sqrt{12}+\sqrt{75}+\sqrt{27}\right):\sqrt{15}\)
\(=\frac{\sqrt{3}\left(2+5+3\right)}{\sqrt{15}}=\frac{10}{\sqrt{5}}=2\sqrt{5}\)
6) Ta có: \(\frac{1}{2}\sqrt{48}-2\sqrt{75}-\frac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\frac{1}{3}}\)
\(=\sqrt{48\cdot\frac{1}{4}}-\sqrt{75\cdot4}-\sqrt{3}+5\sqrt{\frac{4}{3}}\)
\(=\sqrt{12}-\sqrt{300}-\sqrt{3}+\sqrt{25\cdot\frac{4}{3}}\)
\(=\sqrt{12}-\sqrt{300}-\sqrt{3}+\sqrt{\frac{100}{3}}\)
\(=\sqrt{3}\left(2-10-1+\frac{10}{3}\right)\)
\(=-\frac{17\sqrt{3}}{3}=-\frac{17}{\sqrt{3}}\)
27 tháng 9 2020
Bài 1: Thực hiện phép tính
a) Ta có: \(\frac{3+\sqrt{7}}{3-\sqrt{7}}-\frac{3-\sqrt{7}}{3+\sqrt{7}}\)
\(=\frac{\left(3+\sqrt{7}\right)^2}{\left(3-\sqrt{7}\right)\left(3+\sqrt{7}\right)}-\frac{\left(3-\sqrt{7}\right)^2}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}\)
\(=\frac{9+6\sqrt{7}+7-\left(9-6\sqrt{7}+7\right)}{9-7}\)
\(=\frac{16+6\sqrt{7}-16+6\sqrt{7}}{2}\)
\(=\frac{12\sqrt{7}}{2}=6\sqrt{7}\)
b)Sửa đề: \(\left(\frac{\sqrt{2}+5}{\sqrt{2}-5}-\frac{\sqrt{2}-5}{\sqrt{2}+5}\right):\frac{\sqrt{2}}{23}\)
Ta có: \(\left(\frac{\sqrt{2}+5}{\sqrt{2}-5}-\frac{\sqrt{2}-5}{\sqrt{2}+5}\right):\frac{\sqrt{2}}{23}\)
\(=\left(\frac{\left(\sqrt{2}+5\right)^2}{\left(\sqrt{2}-5\right)\left(\sqrt{2}+5\right)}-\frac{\left(\sqrt{2}-5\right)^2}{\left(\sqrt{2}+5\right)\left(\sqrt{2}-5\right)}\right)\cdot\frac{23}{\sqrt{2}}\)
\(=\left(\frac{27+10\sqrt{2}-\left(27-10\sqrt{2}\right)}{2-25}\right)\cdot\frac{23}{\sqrt{2}}\)
\(=\frac{27+10\sqrt{2}-27+10\sqrt{2}}{-23}\cdot\frac{23}{\sqrt{2}}\)
\(=\frac{20\sqrt{2}}{-\sqrt{2}}=-20\)
c) Ta có: \(5\sqrt{\frac{1}{5}}+\frac{1}{2}\sqrt{20}+\sqrt{5}\)
\(=\sqrt{25\cdot\frac{1}{5}}+\frac{1}{2}\cdot2\sqrt{5}+\sqrt{5}\)
\(=\sqrt{5}+\sqrt{5}+\sqrt{5}\)
\(=3\sqrt{5}\)
d) Ta có: \(\sqrt{\frac{1}{2}}+\sqrt{4.5}+12.5\)
\(=\frac{1}{\sqrt{2}}+\frac{3}{\sqrt{2}}+12.5\)
\(=2\sqrt{2}+12.5\)
e) Ta có: \(\frac{1}{2}\sqrt{48}-2\sqrt{75}-\sqrt{54}+5\sqrt{1\frac{1}{3}}\)
\(=\frac{1}{2}\cdot4\sqrt{3}-2\cdot5\sqrt{3}-3\sqrt{6}+5\cdot\sqrt{\frac{4}{3}}\)
\(=2\sqrt{3}-10\sqrt{3}-3\sqrt{6}+\frac{10}{\sqrt{3}}\)
\(=-8\sqrt{3}+\frac{10}{\sqrt{3}}-3\sqrt{6}\)
\(=\frac{-24+10}{\sqrt{3}}-\frac{9\sqrt{2}}{\sqrt{3}}\)
\(=\frac{-14-9\sqrt{2}}{\sqrt{3}}\)
ND
19 tháng 10 2019
a, = \(\frac{\sqrt{15}}{10}\) + \(\frac{\sqrt{15}}{30}\) - \(\frac{2\sqrt{15}}{15}\)
= \(\sqrt{15}\left(\frac{1}{10}+\frac{1}{30}-\frac{2}{15}\right)\)
= \(\sqrt{15}.0\)
= 0
b, = \(\left(\frac{\sqrt{5}+\sqrt{3}}{5-3}+\frac{\sqrt{5}-\sqrt{3}}{5-3}\right).\sqrt{5}\)
= \(\frac{\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}}{2}.\sqrt{5}\)
= \(\frac{2\sqrt{5}}{2}.\sqrt{5}\)
= \(\sqrt{5}.\sqrt{5}\)
= 5
c, = \(\frac{5\sqrt{3}}{\sqrt{15}}+\frac{3\sqrt{5}}{\sqrt{15}}\)
= \(\sqrt{5}+\sqrt{3}\)
d, Mình sửa lại đề bài cho bạn : \(\left(2+\sqrt{5}\right)^2-\left(2-\sqrt{5}\right)^2\)
= \(\left(2+\sqrt{5}-2+\sqrt{5}\right)\left(2+\sqrt{5}+2-\sqrt{5}\right)\)
= \(2\sqrt{5}.4\)
= \(8\sqrt{5}\)
e, = \(\frac{4\sqrt{3}}{3}+15\sqrt{3}-3\sqrt{3}-\frac{20\sqrt{3}}{3}\)
= \(\sqrt{3}.\left(\frac{4}{3}+15-3-\frac{20}{3}\right)\)
= \(\sqrt{3}.\frac{20}{3}\)
= \(\frac{20\sqrt{3}}{3}\)
LT
19 tháng 10 2019
a, √320+√160−2√115320+160−2115
b, (1√5−√3+1√5+√3).√5(15−3+15+3).5
c, (5√3+3√5):√15(53+35):15
d, (2+√5)2−(2+√5)2(2+5)2−(2+5)2
e, 13√48+3√75−√27−10√1131348+375−27−10113
\(\sqrt{85}+\frac{4}{\sqrt{3}}+10\sqrt{3}-\frac{10}{\sqrt{3}}\)=\(\sqrt{85}+10\sqrt{3}-2\sqrt{3}=\sqrt{85}+8\sqrt{3}\)
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