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13 tháng 9 2018
\(=\frac{1}{2}\sqrt{16.3}-2\sqrt{25.3}-\frac{\sqrt{3.11}}{\sqrt{11}}+5\sqrt{\frac{1.3+1}{3}}\)
\(=\frac{1}{2}\sqrt{4^2.3}-2\sqrt{5^2.3}-\frac{\sqrt{3}.\sqrt{11}}{\sqrt{11}}+5\sqrt{\frac{4}{3}}\)
\(=\frac{1}{2}.4\sqrt{3}-2.5\sqrt{3}-\sqrt{3}+5\frac{\sqrt{4}}{\sqrt{3}}\)
\(=\frac{4}{2}\sqrt{3}-10\sqrt{3}-\sqrt{3}+5\frac{\sqrt{4}.\sqrt{4}}{\sqrt{3.}\sqrt{3}}\)
\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}+5\frac{2\sqrt{3}}{3}\)
\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}+10\frac{\sqrt{3}}{3}\)
\(=\left(2-10-1+\frac{10}{3}\right)\sqrt{3}\)
\(=-\frac{17}{3}\)
2 tháng 11 2017
\(\frac{1}{2}\sqrt{48}-2\sqrt{75}-\frac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\frac{1}{3}}\)
=\(\frac{1}{2}\sqrt{3.4^2}-2\sqrt{3.5^2}-\sqrt{\frac{33}{11}}+5\sqrt{\frac{4}{3}}\)
\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}+10\sqrt{\frac{1}{3}}\)
\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}+\frac{10}{3}\sqrt{3}\)
\(=\left(2-10-1+\frac{10}{3}\right)\sqrt{3}\)
\(=\frac{-17}{3}\sqrt{3}\)
\(\sqrt{150}+\sqrt{1,6}.\sqrt{60}+4,5\sqrt{2\frac{2}{3}}-\sqrt{6}\)
\(=\sqrt{6.5^2}+\sqrt{96}+4,5\sqrt{\frac{8}{3}}-\sqrt{6}\)
\(=5\sqrt{6}+\sqrt{6.4^2}+4,5\frac{\sqrt{24}}{3}-\sqrt{6}\)
\(=5\sqrt{6}+4\sqrt{6}+\frac{4,5.2\sqrt{6}}{3}-\sqrt{6}\)
\(=8\sqrt{6}+3\sqrt{6}\)
\(=11\sqrt{6}\)
L
25 tháng 11 2020
Tự hòi tự trl :D ?
\(\frac{1}{2}\sqrt{48}-2\sqrt{75}-\frac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\frac{1}{3}}\)
\(=\frac{1}{2}\sqrt{16.3}-2.5\sqrt{3}-\sqrt{3}-\frac{10}{3}\sqrt{3}\)
\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}-\frac{10}{3}\sqrt{3}\)
\(=-9\sqrt{3}+\frac{10}{3}\sqrt{3}=\left(-9+\frac{10}{3}\right)\sqrt{3}\)
\(=-\frac{17}{3}\sqrt{3}\)
\(\sqrt{150}+\sqrt{1,6}.\sqrt{60}+4,5.\sqrt{2\frac{2}{3}}-\sqrt{6}\)
\(=\sqrt{25.6}+\sqrt{1,6.60}+4,8\sqrt{\frac{8}{3}}-\sqrt{6}\)
\(=5\sqrt{6}+\sqrt{16.6}+4,5.\frac{1}{3}\sqrt{3^2.\frac{4.2}{3}}-\sqrt{6}\)
\(=9\sqrt{6}+3\sqrt{6}-\sqrt{6}=11\sqrt{6}\)
TN
5 tháng 6 2019
\(\frac{1}{2}\sqrt{48}-2\sqrt{75}-\frac{\sqrt{33}}{\sqrt{11}}+5\sqrt{\frac{4}{3}}\)
\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}+\frac{10\sqrt{3}}{3}\)
\(=-9\sqrt{3}+\frac{10\sqrt{3}}{3}\)
\(=\frac{-27\sqrt{3}}{3}+\frac{10\sqrt{3}}{3}\)
\(=\frac{-17\sqrt{3}}{3}\)
\(\frac{1-a\sqrt{a}}{1-\sqrt{a}}\) \(=\frac{1^3-\left(\sqrt{a}\right)^3}{1-\sqrt{a}}\)
\(=\frac{\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}{1-\sqrt{a}}\)
\(=a+\sqrt{a}+1\)
chúc bn học tốt
KN
5 tháng 6 2019
\(\frac{1-a\sqrt{a}}{1-\sqrt{a}}\)
\(=\frac{\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}{1-\sqrt{a}}\)
\(=a+\sqrt{a}+1\)
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: \(\frac{1}{2}\sqrt{48}-2\sqrt{75}-\frac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\frac{1}{3}}\)
\(=\sqrt{48\cdot\frac{1}{4}}-\sqrt{75\cdot4}-\sqrt{3}+5\sqrt{\frac{4}{3}}\)
\(=\sqrt{12}-\sqrt{300}-\sqrt{3}+\sqrt{25\cdot\frac{4}{3}}\)
\(=\sqrt{12}-\sqrt{300}-\sqrt{3}+\sqrt{\frac{100}{3}}\)
\(=\sqrt{3}\left(2-10-1+\frac{10}{3}\right)\)
\(=-\frac{17\sqrt{3}}{3}=-\frac{17}{\sqrt{3}}\)
25 tháng 6 2017
1. \(\frac{1}{2}\sqrt{48}-2\sqrt{75}-\frac{\sqrt{33}}{\sqrt{11}}+\sqrt{84}\)= -6,423305878
2. \(\sqrt{150}+\sqrt{1,6}\sqrt{60}+4,5\sqrt{2\frac{2}{3}}-\sqrt{6}\)= 24,79207036
NHA 
Vũ Hoàng Thiên An ! ! !
K VÀ KB NHA !
9 tháng 10 2017
1) \(2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}\)
\(=2\sqrt{5}-\sqrt{5^2.5}-\sqrt{4^2.5}+\sqrt{11^2.5}\)
\(=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}\)
\(=4\sqrt{5}\)
2) \(\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{15-\sqrt{6^2.6}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{\left(\sqrt{6}\right)^2-6\sqrt{6}+3^2}+\sqrt{\left(2\sqrt{6}\right)^2-12\sqrt{6}+3^2}\)
\(=\sqrt{\left(\sqrt{6}-3\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=\left|\sqrt{6}-3\right|+\left|2\sqrt{6}-3\right|\)
\(=3-\sqrt{6}+2\sqrt{6}-3\) ( vi \(\sqrt{6}-3< 0\))
\(=\sqrt{6}\)
5) \(2\sqrt{\frac{16}{3}}-3\sqrt{\frac{1}{27}}-6\sqrt{\frac{4}{75}}\)
\(=2\frac{4}{\sqrt{3}}-3.\frac{1}{3}-6\sqrt{\frac{2^2}{3.5^2}}\)
\(=\frac{8\sqrt{3}}{3}-1-6.\frac{2}{5}.\sqrt{\frac{1}{3}}\)
\(=8\frac{\sqrt{3}}{3}-1-\frac{12}{5}.\frac{\sqrt{3}}{3}\)
\(=\frac{28}{5}.\frac{\sqrt{3}}{3}-1\)
OM
7 tháng 8 2018
Báo cáo sai phạm
1) 2√5−√125−√80+√605
=2√5−√52.5−√42.5+√112.5
=2√5−5√5−4√5+11√5
=4√5
2) √15−√216+√33−12√6
=√15−√62.6+√33−12√6
=√15−6√6+√33−12√6
=√(√6)2−6√6+32+√(2√6)2−12√6+32
=√(√6−3)2+√(2√6−3)2
=|√6−3|+|2√6−3|
=3−√6+2√6−3 ( vi √6−3<0)
=√6
5) 2√163 −3√127 −6√475
=24√3 −3.13 −6√223.52
=8√33 −1−6.25 .√13
=8√33 −1−125 .√33
=285 .√33 −1
\(=\frac{1}{2}\sqrt{48}-2\sqrt{75}-\frac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\frac{1}{3}}\)
\(=\frac{1}{2}4\sqrt{3}-2.5\sqrt{3}-\sqrt{3}+\frac{10}{\sqrt{3}}=-9\sqrt{3}+\frac{10}{\sqrt{3}}=\frac{-17\sqrt{3}}{3}\)