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1 tháng 10 2018
\(a)\)\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=\)\(\sqrt{6-6\sqrt{6}+9}+\sqrt{24-12\sqrt{6}+9}\)
\(=\)\(\sqrt{\left(\sqrt{6}+3\right)}+\sqrt{\left(\sqrt{24}+3\right)}\)
\(=\)\(\left|\sqrt{6}+3\right|+\left|\sqrt{24}+3\right|\)
\(=\)\(\sqrt{6}+3+\sqrt{24}+3\)
\(=\)\(\sqrt{6}\left(1+\sqrt{4}\right)+9\)
\(=\)\(3\sqrt{6}+9\)
Chúc bạn học tốt ~
1 tháng 10 2018
\(b)\)\(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\)
\(=\)\(\left|2-\sqrt{3}\right|+\sqrt{3-2\sqrt{3}+1}\)
\(=\)\(2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}\) ( vì \(2=\sqrt{4}>\sqrt{3}\) )
\(=\)\(2-\sqrt{3}+\left|\sqrt{3}-1\right|\)
\(=\)\(2-\sqrt{3}+\sqrt{3}-1\) ( vì \(\sqrt{3}>\sqrt{1}=1\) )
\(=\)\(1\)
Chúc bạn học tốt ~
PS : mới lớp 8 sai thì thông cảm >.<
NW
28 tháng 6 2016
mình ghi nhầm pn ơi.. bài 2 là \(\left(3-\sqrt{2}\right)\cdot\sqrt{11+6\sqrt{6}}\)
23 tháng 8 2019
32, \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
=\(\sqrt{9-2.3.\sqrt{6}+6}+\sqrt{33-2.3.2\sqrt{6}}\)
=\(\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{24-2.3.2\sqrt{6}+9}\)
=\(\left|3-\sqrt{6}\right|+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
=\(3-\sqrt{6}+\left|2\sqrt{6}-3\right|\)=\(3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)
33, \(\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}=\sqrt{\left(\sqrt{5}-1\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}=\left|\sqrt{5}-1\right|+\sqrt{5}+1=\sqrt{5}-1+\sqrt{5}+1=2\sqrt{5}\)
34, \(\sqrt{8-2\sqrt{15}}-\sqrt{23-4\sqrt{15}}\)
=\(\sqrt{8-2.\sqrt{3}.\sqrt{5}}-\sqrt{23-2.2.\sqrt{5}.\sqrt{3}}\)
=\(\sqrt{5-2\sqrt{3}.\sqrt{5}+3}-\sqrt{\left(2\sqrt{5}\right)^2-2.2\sqrt{5}.\sqrt{3}+3}\)
=\(\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{\left(2\sqrt{5}-\sqrt{3}\right)^2}\)
=\(\left|\sqrt{5}-\sqrt{3}\right|-\left|2\sqrt{5}-\sqrt{3}\right|=\sqrt{5}-\sqrt{3}-2\sqrt{5}+\sqrt{3}=-\sqrt{5}\)
23 tháng 8 2019
35,\(\sqrt{31-8\sqrt{15}}+\sqrt{24-6\sqrt{15}}\)
=\(\sqrt{16-2.4.\sqrt{15}+15}+\sqrt{15-2.3.\sqrt{15}+9}\)
=\(\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}-3\right)^2}\)
=\(\left|4-\sqrt{15}\right|+\left|\sqrt{15}-3\right|\)
=\(4-\sqrt{15}+\sqrt{15}-3\)
=1
36, \(\sqrt{49-5\sqrt{96}}-\sqrt{49+5\sqrt{96}}\)
=\(\sqrt{49-2.5.\sqrt{24}}-\sqrt{49+2.5\sqrt{24}}=\sqrt{25-2.5.\sqrt{24}+24}-\sqrt{25+2.5.\sqrt{24}+24}=\sqrt{\left(5-\sqrt{24}\right)^2}-\sqrt{\left(5+\sqrt{24}\right)^2}\)
=\(\left|5-\sqrt{24}\right|-\left|5+\sqrt{24}\right|=5-\sqrt{24}-5-\sqrt{24}=-2\sqrt{24}\)
37, \(\sqrt{3+2\sqrt{2}}+\sqrt{5-2\sqrt{6}}=\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
=\(\left|\sqrt{2}+1\right|+\left|\sqrt{3}-\sqrt{2}\right|=\sqrt{2}+1+\sqrt{3}-\sqrt{2}=\sqrt{3}+1\)
4 tháng 9 2016
1. \(\sqrt{7+2\sqrt{10}}-\sqrt{7-2\sqrt{10}}=\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}\\ =\sqrt{5}+\sqrt{2}-\sqrt{5}+\sqrt{2}=2\sqrt{2}\)
2. \(\sqrt{12-6\sqrt{3}}+\sqrt{21-12\sqrt{3}}=\sqrt{\left(3-\sqrt{3}\right)^2}+\sqrt{\left(2\sqrt{3}-3\right)^2}\\ =3-\sqrt{3}+2\sqrt{3}-3=\sqrt{3}\)
3. \(\sqrt{33-12\sqrt{6}}+\sqrt{15-6\sqrt{6}}=\sqrt{\left(2\sqrt{6}-3\right)^2}+\sqrt{\left(3+\sqrt{6}\right)^2}\\ =2\sqrt{6}-3+3+\sqrt{6}=3\sqrt{6}\)
4 tháng 9 2016
1.\(\sqrt{7+2\sqrt{10}}-\sqrt{7-2\sqrt{10}}=\sqrt{\left(\sqrt{2}+\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}\)
\(=\sqrt{5}+\sqrt{2}-\left(\sqrt{5}-\sqrt{2}\right)=2\sqrt{2}\)
2. \(\sqrt{12-6\sqrt{3}+\sqrt{21-12\sqrt{3}}}=\sqrt{12-6\sqrt{3}+\sqrt{\left(3-2\sqrt{3}\right)^2}}\)
\(=\sqrt{12-6\sqrt{3}+2\sqrt{3}-3}=\sqrt{9-4\sqrt{3}}\)
3. \(\sqrt{33-12\sqrt{6}}+\sqrt{15-6\sqrt{6}}=\sqrt{\left(2\sqrt{6}-3\right)^2}+\sqrt{\left(\sqrt{6}-3\right)^2}\)
\(=2\sqrt{6}-3+3-\sqrt{6}=\sqrt{6}\)
26 tháng 7 2019
\( 2)2\sqrt {\dfrac{{16}}{3}} - 3\sqrt {\dfrac{1}{{27}}} - 6\sqrt {\dfrac{4}{{75}}} \\ = 2.\dfrac{4}{{\sqrt 3 }} - 3.\dfrac{1}{{\sqrt {27} }} - 6\dfrac{2}{{\sqrt {75} }}\\ = \dfrac{8}{{\sqrt 3 }} - \dfrac{3}{{3\sqrt 3 }} - \dfrac{{12}}{{5\sqrt 2 }}\\ = \dfrac{{8\sqrt 3 }}{3} - \dfrac{{\sqrt 3 }}{3} - \dfrac{{4\sqrt 3 }}{5}\\ = \dfrac{{23\sqrt 3 }}{{15}}\\ 3)2\sqrt {27} - 6\sqrt {\dfrac{4}{3}} + \dfrac{3}{5}\sqrt {75} \\ = 6\sqrt 3 - \dfrac{{12}}{{\sqrt 3 }} + 3\sqrt 3 \\ = 9\sqrt 3 - 4\sqrt 3 \\ = 5\sqrt 3 \)
DK
1
29 tháng 6 2017
\(\sqrt{15+6\sqrt{6}}-\sqrt{33-12\sqrt{6}}\)
\(\Leftrightarrow\sqrt{\left(3+\sqrt{6}\right)^2}-\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(\Leftrightarrow3+\sqrt{6}-2\sqrt{6}+3\)
\(\Leftrightarrow6-\sqrt{6}\)
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
12 tháng 10 2019
a) \(\left(\sqrt{27}-\sqrt{12}-\sqrt{108}-\sqrt{192}\right):\sqrt{3}=\left(3\sqrt{3}-2\sqrt{3}-6\sqrt{3}-8\sqrt{3}\right):\sqrt{3}=\left(-13\sqrt{3}\right):\sqrt{3}=-13\sqrt{3}.\frac{1}{\sqrt{3}}=-13\)
c) \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}=\left|3-\sqrt{6}\right|+\left|2\sqrt{6}-3\right|=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)
VH
12 tháng 10 2019
a, \(=\left(3\sqrt{3}-2\sqrt{3}-6\sqrt{3}-8\sqrt{3}\right):\sqrt{3}\)
\(=\frac{-13\sqrt{3}}{\sqrt{3}}=-13\)
b, \(=\frac{\sqrt{2}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}.\frac{3+2\sqrt{7}}{1+\sqrt{3}}\)
\(=\frac{\sqrt{2}\left(3+2\sqrt{7}\right)}{1+\sqrt{3}}\)
c, \(=\sqrt{6-6\sqrt{6} +9}+\sqrt{24-2.2\sqrt{6}.3+9}\)
\(=\sqrt{\left(\sqrt{6}-3\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)
\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)