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Trả lời:
\(\frac{2\sqrt{5}-5\sqrt{2}}{\sqrt{2}-\sqrt{5}}+\frac{6}{2-\sqrt{10}}+\sqrt{67+12\sqrt{7}}\)
\(=\frac{\sqrt{2}.\sqrt{5}.\left(\sqrt{2}-\sqrt{5}\right)}{\sqrt{2}-\sqrt{5}}-\frac{6}{\sqrt{10}-2}+\sqrt{63+12\sqrt{7}+4}\)
\(=\sqrt{2}.\sqrt{5}-\frac{6.\left(\sqrt{10}+2\right)}{10-4}+\sqrt{\left(3\sqrt{7}+2\right)^2}\)
\(=\sqrt{10}-\sqrt{10}-2+3\sqrt{7}+2\)
\(=3\sqrt{7}\)
Bài làm:
\(\frac{\sqrt{5}+2}{\sqrt{5}-2}=\frac{\left(\sqrt{5}+2\right)^2}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}=\frac{5+2.2.\sqrt{5}+4}{5-4}\)
\(=9+4\sqrt{5}\)
Học tốt!!!!
\(\frac{\sqrt{5}+2}{\sqrt{5}-2}=\frac{5+2\sqrt{5}+4+2\sqrt{5}}{\sqrt{5}^2-2^2}\)
\(=\frac{9+4\sqrt{5}}{5-4}=9+4\sqrt{5}\)
@Học tốt@
Bài làm:
Ta có: \(\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}=\frac{\left(\sqrt{7}-\sqrt{5}\right)^2}{\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)}=\frac{7-2.\sqrt{7}.\sqrt{5}-5}{7-5}\)
\(=\frac{2-2\sqrt{35}}{2}=1-\sqrt{35}\)
Học tốT!!!!
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) \(2\sqrt{\frac{16}{3}}-3\sqrt{\frac{1}{27}}-6\sqrt{\frac{4}{75}}\)
\(=2\frac{4}{\sqrt{3}}-3.\frac{1}{3}-6\sqrt{\frac{2^2}{3.5^2}}\)
\(=\frac{8\sqrt{3}}{3}-1-6.\frac{2}{5}.\sqrt{\frac{1}{3}}\)
\(=8\frac{\sqrt{3}}{3}-1-\frac{12}{5}.\frac{\sqrt{3}}{3}\)
\(=\frac{28}{5}.\frac{\sqrt{3}}{3}-1\)
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
Trả lời:
\(\frac{2}{\sqrt{5}+\sqrt{3}}-\sqrt{\frac{2}{4-\sqrt{15}}}+6\sqrt{\frac{1}{3}}\)
\(=\frac{2.\left(\sqrt{5}-\sqrt{3}\right)}{5-3}-\sqrt{\frac{2\times2}{2\times\left(4-\sqrt{15}\right)}}+6\times\frac{1}{\sqrt{3}}\)
\(=\frac{2.\left(\sqrt{5}-\sqrt{3}\right)}{2}-\sqrt{\frac{4}{8-2\sqrt{15}}}+6\times\frac{\sqrt{3}}{3}\)
\(=\sqrt{5}-\sqrt{3}-\sqrt{\frac{4}{5-2\sqrt{15}+3}}+2\sqrt{3}\)
\(=\sqrt{5}-\sqrt{3}-\sqrt{\frac{4}{\left(\sqrt{5}-\sqrt{3}\right)^2}}+2\sqrt{3}\)
\(=\sqrt{5}-\sqrt{3}-\frac{2}{\sqrt{5}-\sqrt{3}}+2\sqrt{3}\)
\(=\sqrt{5}+\sqrt{3}-\frac{2}{\sqrt{5}-\sqrt{3}}\)
\(=\frac{\left(\sqrt{5}-\sqrt{3}\right).\left(\sqrt{5}+3\right)-2}{\sqrt{5}-\sqrt{3}}\)
\(=\frac{5-3-2}{\sqrt{5}-\sqrt{3}}\)
\(=0\)
Học tốt