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7 tháng 9 2020
+) Ta có: \(2\sqrt{75}-4\sqrt{27}+3\sqrt{12}\)
\(=2\sqrt{25}.\sqrt{3}-4\sqrt{9}.\sqrt{3}+3\sqrt{4}.\sqrt{3}\)
\(=10.\sqrt{3}-12.\sqrt{3}+6.\sqrt{3}\)
\(=4\sqrt{3}\approx6,9282\)
+) Ta có:\(\sqrt{x+6\sqrt{x-9}}\)
\(=\sqrt{x-9+6\sqrt{x-9}+9}\)
\(=\sqrt{\left(\sqrt{x-9}-3\right)^2}\)
\(=\left|\sqrt{x-9}-3\right|\)
7 tháng 9 2020
\(\frac{2}{\sqrt{5}+\sqrt{3}}+\frac{1}{2-\sqrt{3}}=\frac{2\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}+\frac{2+\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\)
\(=\frac{2\left(\sqrt{5}-\sqrt{3}\right)}{5-3}+\frac{2+\sqrt{3}}{4-3}=\sqrt{5}-\sqrt{3}+2+\sqrt{3}=\sqrt{5}+2\)
NT
0
10 tháng 12 2015
\(2\sqrt{9\left(x-3\right)}-\sqrt{4\left(x-3\right)}=10+\frac{1}{2}\)
\(6\sqrt{\left(x-3\right)}-2\sqrt{\left(x-3\right)}=\frac{21}{2}\)
\(4\sqrt{\left(x-3\right)}=\frac{21}{2}\)
\(\sqrt{\left(x-3\right)}=\frac{21}{8}\)
\(x-3=\frac{441}{64}\)
\(x=\frac{633}{64}\)
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
11 tháng 8 2023
\(=12\sqrt{3}-2\cdot4\sqrt{3}+3\cdot2\sqrt{3}-5\sqrt{3}\)
\(=12\sqrt{3}-8\sqrt{3}+6\sqrt{3}-5\sqrt{3}\)
\(=5\sqrt{3}\)
HK
0
27 tháng 12 2017
a)
\(7\sqrt{12}+\frac{1}{3}\sqrt{27}-\sqrt{75}\)
\(=14\sqrt{3}+\sqrt{3}-5\sqrt{3}\)
\(=10\sqrt{3}\)
b)
\(\left(2\sqrt{20}+\sqrt{125}-3\sqrt{80}\right):5\)
\(=\left(4\sqrt{5}+5\sqrt{5}-12\sqrt{5}\right):5\)
\(=-3\sqrt{5}:5\)
\(=\frac{-3\sqrt{5}}{5}\)
c)
\(3\sqrt{12a}-5\sqrt{3a}+\sqrt{48a}\)
\(=6\sqrt{3a}-5\sqrt{3a}+4\sqrt{3a}\)
\(=5\sqrt{3a}\)
`sqrt{12}-sqrt{75}+sqrt{27}`
`=sqrt{4.3}-sqrt{25.3}+sqrt{9.3}`
`=2sqrt3-5sqrt3+3sqrt3=0`
Ta có : \(\sqrt{12}-\sqrt{75}+\sqrt{27}=2\sqrt{3}-5\sqrt{3}+3\sqrt{3}=0\)