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AH
Akai Haruma
Giáo viên
4 tháng 7 2019
Lời giải:
d)
\(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{6+2\sqrt{5-\sqrt{13+2\sqrt{12}}}}\)
\(=\sqrt{6+2\sqrt{5-\sqrt{(\sqrt{12}+1)^2}}}=\sqrt{6+2\sqrt{5-(\sqrt{12}+1)}}\)
\(=\sqrt{6+2\sqrt{4-2\sqrt{3}}}=\sqrt{6+2\sqrt{(\sqrt{3}-1)^2}}\)
\(=\sqrt{6+2(\sqrt{3}-1)}=\sqrt{4+2\sqrt{3}}=\sqrt{(\sqrt{3}+1)^2}\)
\(=\sqrt{3}+1\)
e)
\(\frac{2}{\sqrt{3}-1}-\frac{3-2\sqrt{3}}{2-\sqrt{3}}=\frac{2}{\sqrt{3}-1}+\frac{\sqrt{3}(2-\sqrt{3})}{2-\sqrt{3}}=\frac{2}{\sqrt{3}-1}+\sqrt{3}\)
\(=\frac{2(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}+\sqrt{3}=\frac{2(\sqrt{3}+1)}{3-1}+\sqrt{3}=\sqrt{3}+1+\sqrt{3}=2\sqrt{3}+1\)
Y
4 tháng 7 2019
d)
\(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\\ =\sqrt{6+2\sqrt{5-\sqrt{12+2\cdot\sqrt{12}\cdot1+1}}}\\ =\sqrt{6+2\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}\\ =\sqrt{6+2\sqrt{5-\sqrt{12}-1}}\\ =\sqrt{6+2\sqrt{4-\sqrt{12}}}\\ =\sqrt{6+2\sqrt{3-2\cdot\sqrt{3}\cdot1+1}}\\ =\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\\ =\sqrt{6+2\left(\sqrt{3}-1\right)}\\ =\sqrt{6-2\sqrt{3}-2}\\ =\sqrt{4-2\sqrt{3}}\\ =\sqrt{3-2\sqrt{3}+1}\\ =\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
e)
\(\frac{2}{\sqrt{3}-1}-\frac{3-2\sqrt{3}}{2-\sqrt{3}}\\ =\frac{2}{\sqrt{3}-1}+\frac{\sqrt{3}\left(\sqrt{3}-2\right)}{\sqrt{3}-2}\\ =\frac{2}{\sqrt{3}-1}+\sqrt{3}\\ =\frac{2+\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\\ =\frac{5-\sqrt{3}}{\sqrt{3}-1}\\ =\frac{\left(5-\sqrt{3}\right)\left(\sqrt{3}-1\right)}{\left(\sqrt{3}\right)^2-1}\\ -\frac{6\sqrt{3}-8}{2}=3\sqrt{3}-4\)
(bạn nhớ ktr đã nha)
8 tháng 5 2019
a, \(\sqrt{2}A=\sqrt{10-2\sqrt{3.7}}+\sqrt{10+2\sqrt{3.7}}\)
\(=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{3}+\sqrt{7}\right)^2}\)
\(=\left|\sqrt{7}-\sqrt{3}\right|+\left|\sqrt{7}+\sqrt{3}\right|\)
\(=\sqrt{7}-\sqrt{3}+\sqrt{3}+\sqrt{7}=2\sqrt{7}\)
\(\Rightarrow A=\sqrt{14}\)
b, \(B=\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+\frac{\sqrt{5}\left(\sqrt{5}-2\right)}{2\left(\sqrt{5}-2\right)}\)
\(=\sqrt{5}+\frac{\sqrt{5}}{2}=\frac{3\sqrt{5}}{2}\)
c, \(C=\left(1-\sqrt{11}\right)\left(\sqrt{11}+1\right)=1-11=-10\)
d, \(D=\frac{\sqrt{2}\left(\sqrt{2}+\sqrt{3}\right)}{2-3}-\frac{\sqrt{2}\left(\sqrt{2}-\sqrt{3}\right)}{2-3}\)
\(=-2-\sqrt{6}+2-\sqrt{6}=-2\sqrt{6}\)
8 tháng 5 2019
d) \(\frac{1}{\sqrt{3}-\sqrt{5}}-\frac{1}{\sqrt{3}+\sqrt{5}}=\frac{\sqrt{3}+\sqrt{5}}{\left(\sqrt{3}-\sqrt{5}\right)\left(\sqrt{3}+\sqrt{5}\right)}-\frac{\sqrt{3}-\sqrt{5}}{\left(\sqrt{3}-\sqrt{5}\right)\left(\sqrt{3}+\sqrt{5}\right)}=\frac{\sqrt{3}+\sqrt{5}-\sqrt{3}+\sqrt{5}}{\left(\sqrt{3}-\sqrt{5}\right)\left(\sqrt{3}+\sqrt{5}\right)}=\frac{2\sqrt{5}}{3-5}=\frac{2\sqrt{5}}{-2}=-\sqrt{5}\)c) \(\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}}=\sqrt{3+2\sqrt{3}.\sqrt{2}+2}+\sqrt{3-2\sqrt{3}.\sqrt{2}+2}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}\)
b) \(\sqrt{9+4\sqrt{5}}+\sqrt{9-4\sqrt{5}}=\sqrt{5+2.\sqrt{5}.2+4}+\sqrt{5-2.\sqrt{5}.2+4}=\sqrt{\left(\sqrt{5}+2\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}=\sqrt{5}+2+\sqrt{5}-2=2\sqrt{5}\)a) \(\sqrt{27}+\sqrt{243}-6\sqrt{12}=\sqrt{9.3}+\sqrt{81.3}-6\sqrt{4.3}=3\sqrt{3}+9\sqrt{3}-12\sqrt{3}=0\)
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
PN
29 tháng 7 2020
\(A=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=1+\sqrt{2}\)
AH
Akai Haruma
Giáo viên
4 tháng 7 2019
Lời giải:
\(B=4\sqrt{20}+\sqrt{6-2\sqrt{5}}-15\sqrt{\frac{1}{5}}\)
\(=4\sqrt{4}.\sqrt{5}+\sqrt{5-2\sqrt{5}+1}-3.\sqrt{25}.\sqrt{\frac{1}{5}}\)
\(=8\sqrt{5}+\sqrt{(\sqrt{5}-1)^2}-3\sqrt{5}\)
\(=(8-3)\sqrt{5}+(\sqrt{5}-1)=5\sqrt{5}+\sqrt{5}-1=6\sqrt{5}-1\)