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

a)\(2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}\)

= \(2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}\)

= \(4\sqrt{5}\)

b) \(\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}\)

= \(\sqrt{3\left(5-2\sqrt{6}\right)}-\sqrt{33-12\sqrt{6}}\)

= \(\sqrt{3\left(5-2\sqrt{6}\right)}-\sqrt{3\left(11-4\sqrt{6}\right)}\)

\(a,2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}\)

\(=2\sqrt{5}-\sqrt{5^2.5}-\sqrt{4^2.5}+\sqrt{11^2.5}\)

\(=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}\)

\(=4\sqrt{5}\)

\(b,\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}\)

\(=\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)

\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(3-2\sqrt{6}\right)^2}\)

\(=|3-\sqrt{6}|+|3-2\sqrt{6}|\)

\(=3-\sqrt{6}+2\sqrt{6}-3\)

\(=\sqrt{6}\)

a) Kết quả rút gọn xấu (+dài) nữa. (có thể đề sai)

b) 

\(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}\)

\(=\left[\frac{-\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\frac{-\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right].\left(\sqrt{7}-\sqrt{5}\right)\)

\(=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)=-\left(7-5\right)=-2\)

c) \(\frac{\sqrt{5-2\sqrt{6}}+\sqrt{8-2\sqrt{15}}}{\sqrt{7+2\sqrt{10}}}=\frac{\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}}{\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}}\)

\(=\frac{\sqrt{3}-\sqrt{2}+\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{2}}=\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}=\frac{\left(\sqrt{5}-\sqrt{2}\right)^2}{3}\)

a) \(\left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right).\frac{1}{\sqrt{6}}=\left[\frac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-2\sqrt{6}\right].\frac{1}{\sqrt{6}}\)

\(=\left(\frac{\sqrt{6}}{2}-2\sqrt{6}\right).\frac{1}{\sqrt{6}}=\frac{1}{2}-2=-\frac{3}{2}\)

Mysterious Person giúp e với! Em cảm ơn!!!

a: \(=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}=4\sqrt{5}\)

b: \(=2\sqrt{5}-2-2\sqrt{5}=-2\)

c: \(=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)

d: \(=\dfrac{2\left(2\sqrt{2}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{3}-2\sqrt{2}\right)}-\dfrac{1}{\sqrt{6}}\)

\(=\dfrac{-3}{\sqrt{6}}=-\dfrac{3\sqrt{6}}{6}=-\dfrac{\sqrt{6}}{2}\)

e: \(=\dfrac{8}{3}\sqrt{3}-\dfrac{1}{3}\sqrt{3}-\dfrac{4}{5}\sqrt{3}=\dfrac{23}{15}\sqrt{3}\)

Thêm câu này hộ tớ nx nhé !
e) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right).\left(\sqrt{2}-3\sqrt{0.4}\right)\)

\(a,\left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right)\cdot\frac{1}{\sqrt{6}}\)

\(=\left(\frac{\sqrt{12}-\sqrt{6}}{2\left(\sqrt{2}-1\right)}-\frac{6\sqrt{6}}{3}\right)\cdot\frac{1}{\sqrt{6}}\)

\(=\left(\frac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-2\sqrt{6}\right)\cdot\frac{1}{\sqrt{6}}\)

\(=\left(\frac{\sqrt{6}}{2}-\frac{4\sqrt{6}}{2}\right)\cdot\frac{1}{\sqrt{6}}\)

\(=\frac{\sqrt{6}-4\sqrt{6}}{2}\cdot\frac{1}{\sqrt{6}}\)

\(=\frac{-3\sqrt{6}}{2}\cdot\frac{1}{\sqrt{6}}\)

\(=-\frac{3}{2}\)

\(b,\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}\)

\(=\left(\frac{\sqrt{7}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}+\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}\right).\left(\sqrt{7}-\sqrt{5}\right)\)

\(=\left(\frac{-\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}-\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right).\left(\sqrt{7}-\sqrt{5}\right)\)

\(=\left(-\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)

\(=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)

\(=-\left(7-5\right)\)

\(=-2\)

Giúp mình :<

a) \(\frac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}\)

= \(\frac{\left(10+2\sqrt{10}\right)\left(1-\sqrt{5}\right)}{\left(\sqrt{5}+\sqrt{2}\right)\left(1-\sqrt{5}\right)}+\frac{8\left(\sqrt{5}+\sqrt{2}\right)}{\left(1-\sqrt{5}\right)\left(\sqrt{5}+\sqrt{2}\right)}\)

= \(\frac{\left(10+2\sqrt{10}\right)\left(1-\sqrt{5}\right)+8\left(\sqrt{5}+\sqrt{2}\right)}{\left(1-\sqrt{5}\right)\left(\sqrt{5}+\sqrt{2}\right)}\)

= \(\frac{10-2\sqrt{5}+2\sqrt{10}-2\sqrt{2}}{\sqrt{5}+\sqrt{2}-5-\sqrt{10}}\)

= \(\frac{2\left(5-\sqrt{5}+\sqrt{10}-\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}-5-\sqrt{10}}\)

= -2

b); c); d) làm tương tự

a) \(\dfrac{1}{\sqrt{3}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{\sqrt{3}}\sqrt{\dfrac{5}{12}-\dfrac{1}{\sqrt{6}}}\)

\(=\dfrac{\sqrt{3}}{3}+\dfrac{\sqrt{2}}{6}+\dfrac{1}{\sqrt{3}}\sqrt{\dfrac{5}{12}-\dfrac{\sqrt{6}}{6}}\)

\(=\dfrac{\sqrt{3}}{3}+\dfrac{\sqrt{2}}{6}+\dfrac{\sqrt{\dfrac{5}{12}-\dfrac{\sqrt{6}}{6}}}{\sqrt{3}}\)

\(=\dfrac{\sqrt{3}}{3}+\dfrac{\sqrt{2}}{6}+\dfrac{\sqrt{\left(\dfrac{5}{12}-\dfrac{\sqrt{6}}{6}\right)\cdot3}}{3}\)

\(=\dfrac{\sqrt{3}}{3}+\dfrac{\sqrt{2}}{6}+\dfrac{\sqrt{\dfrac{5}{4}-\dfrac{\sqrt{6}}{2}}}{3}\)

\(=\dfrac{\sqrt{3}+\sqrt{\dfrac{5}{4}-\dfrac{\sqrt{6}}{2}}}{3}+\dfrac{\sqrt{2}}{6}\)

b) \(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}=...\)

c) \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}=...\)

d) \(\dfrac{2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}\)

\(=\dfrac{2\sqrt{3-\sqrt{3+\sqrt{13+4\sqrt{3}}}}}{\sqrt{6}-\sqrt{2}}\)

\(=\dfrac{2\sqrt{3-\sqrt{3+\sqrt{\left(1+2\sqrt{3}\right)^2}}}}{\sqrt{6}-\sqrt{2}}\)

\(=\dfrac{2\sqrt{3-\sqrt{3+1+2\sqrt{3}}}}{\sqrt{6}-\sqrt{2}}\)

\(=\dfrac{2\sqrt{3-\sqrt{3+2\sqrt{3}+1}}}{\sqrt{6}-\sqrt{2}}\)

\(=\dfrac{2\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}{\sqrt{6}-\sqrt{2}}\)

\(=\dfrac{2\sqrt{3-\left(\sqrt{3}+1\right)}}{\sqrt{6}-\sqrt{2}}\)

\(=\dfrac{2\sqrt{3-\left(\sqrt{3}+1\right)}\cdot\left(\sqrt{6}+\sqrt{2}\right)}{4}\)

\(=\dfrac{\sqrt{3\left(\sqrt{3}+1\right)}\cdot\left(\sqrt{6}+\sqrt{2}\right)}{2}\)

\(=\dfrac{\sqrt{3-\sqrt{3}-1}\sqrt{\left(\sqrt{6}+\sqrt{2}\right)^2}}{2}\)

\(=\dfrac{\sqrt{\left(3-\sqrt{3}-1\right)\cdot\left(\sqrt{6}+\sqrt{2}\right)^2}}{2}\)

\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\cdot\left(6+2\sqrt{12}+2\right)}}{2}\)

\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\cdot\left(6+4\sqrt{3}+2\right)}}{2}\)

\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\cdot\left(8+4\sqrt{3}\right)}}{2}\)

\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\cdot4\left(2+\sqrt{3}\right)}}{2}\)

\(=\dfrac{\sqrt{\left(4-3\right)\cdot4}}{2}\)

\(=\dfrac{\sqrt{1\cdot4}}{2}\)

\(=\dfrac{2}{2}\)

\(=1\)

a. \(\frac{26}{5-2\sqrt{3}}\)=\(\frac{26\cdot\left(5+2\sqrt{3}\right)}{\left(5-2\sqrt{3}\right)\left(5+2\sqrt{3}\right)}\)=\(\frac{26\cdot\left(5+2\sqrt{3}\right)}{5^2-\left(2\sqrt{3}\right)^2}=\frac{26\cdot\left(5+2\sqrt{3}\right)}{13}=2\cdot\left(5+2\sqrt{3}\right)=10+4\sqrt{3}\)

b.\(\frac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}=\frac{\sqrt{3}\cdot\left(3\sqrt{3}-2\right)}{\sqrt{2}\cdot\left(3\sqrt{3}-2\right)}=\frac{\sqrt{3}}{\sqrt{2}}=\frac{\sqrt{6}}{2}\)

c.\(\frac{2\sqrt{10}-5}{4-\sqrt{10}}=\frac{\sqrt{5}\cdot\left(2\sqrt{2}-\sqrt{5}\right)}{\sqrt{2}\cdot\left(2\sqrt{2}-\sqrt{5}\right)}=\frac{\sqrt{5}}{\sqrt{2}}=\frac{\sqrt{10}}{2}\)

d.\(2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}\)=\(4\sqrt{5}\)