1,

\(2\sqrt{5}-\sqrt{125}-\sqrt{80}\\ =2\sqrt{5}-\sqrt{25\cdot5}-\sqrt{16\cdot5}\\ =2\sqrt{5}-5\sqrt{5}-4\sqrt{5}\\ =-7\sqrt{5}\)

2,

\(3\sqrt{2}-\sqrt{8}+\sqrt{50}-4\sqrt{32}\\ =3\sqrt{2}-\sqrt{4\cdot2}+\sqrt{25\cdot2}-4\sqrt{16\cdot2}\\ =3\sqrt{2}-2\sqrt{2}+5\sqrt{2}-16\sqrt{2}\\=-10\sqrt{2}\)

3,

\(\sqrt{18}-3\sqrt{80}-2\sqrt{50}+2\sqrt{45}\\ =\sqrt{9\cdot2}-3\sqrt{16\cdot5}-2\sqrt{25\cdot2}+2\sqrt{9\cdot5}\\ =3\sqrt{2}-12\sqrt{5}-10\sqrt{2}+6\sqrt{5}\\ =-7\sqrt{2}-6\sqrt{5}\)

4,

\(\sqrt{27}-2\sqrt{3}+2\sqrt{48}-3\sqrt{75}\\ =\sqrt{9\cdot3}-2\sqrt{3}+2\sqrt{16\cdot3}-3\sqrt{25\cdot2}\\ =3\sqrt{3}-2\sqrt{3}+8\sqrt{3}-15\sqrt{3}\\ =-6\sqrt{3}\)

5,

\(3\sqrt{2}-4\sqrt{18}+\sqrt{32}-\sqrt{50}\\ =3\sqrt{2}-4\sqrt{9\cdot2}+\sqrt{16\cdot2}-\sqrt{25\cdot2}\\ =3\sqrt{2}-12\sqrt{2}+4\sqrt{2}-5\sqrt{2}\\ =-10\sqrt{2}\)

6,

\(2\sqrt{3}-\sqrt{75}+2\sqrt{12}-\sqrt{147}\\ =2\sqrt{3}-\sqrt{25\cdot3}+2\sqrt{4\cdot3}-\sqrt{49\cdot3}\\ =2\sqrt{3}-5\sqrt{3}+4\sqrt{3}-7\sqrt{3}\\ =-6\sqrt{3}\)

7,

\(\sqrt{20}-2\sqrt{45}-3\sqrt{80}+\sqrt{125}\\ =\sqrt{4\cdot5}-2\sqrt{9\cdot5}-3\sqrt{16\cdot5}+\sqrt{25\cdot5}\\ =2\sqrt{5}-6\sqrt{5}-12\sqrt{5}+5\sqrt{5}\\ =-11\sqrt{5}\)

8,

\(6\sqrt{12}-\sqrt{20}-2\sqrt{27}+\sqrt{125}\\ =6\sqrt{4\cdot3}-\sqrt{4\cdot5}-2\sqrt{9\cdot3}+\sqrt{25\cdot5}\\ =12\sqrt{3}-2\sqrt{5}-6\sqrt{3}+5\sqrt{5}\\ =6\sqrt{3}+3\sqrt{5}\\ =3\left(2\sqrt{3}+\sqrt{5}\right)\)

9,

\(4\sqrt{24}-2\sqrt{54}+3\sqrt{6}-\sqrt{150}\\ =4\sqrt{4\cdot6}-2\sqrt{9\cdot6}+3\sqrt{6}-\sqrt{25\cdot6}\\ =8\sqrt{6}-6\sqrt{6}+3\sqrt{6}-5\sqrt{6}=0\)

10,

\(2\sqrt{18}-3\sqrt{80}-5\sqrt{147}+5\sqrt{245}-3\sqrt{98}\\ =2\sqrt{9\cdot2}-3\sqrt{16\cdot5}-5\sqrt{49\cdot3}+5\sqrt{49\cdot5}-3\sqrt{49\cdot2}\\ =6\sqrt{2}-12\sqrt{5}-35\sqrt{3}+35\sqrt{5}-21\sqrt{2}\\ =-15\sqrt{2}-35\sqrt{3}+23\sqrt{5}\)

Bài 1:

a) Sửa đề: \(\left(\sqrt{12}+3\sqrt{5}-4\sqrt{135}\right)\cdot\sqrt{3}\)

Ta có: \(\left(\sqrt{12}+3\sqrt{5}-4\sqrt{135}\right)\cdot\sqrt{3}\)

\(=\sqrt{12}\cdot\sqrt{3}+3\sqrt{5}\cdot\sqrt{3}-4\sqrt{135}\cdot\sqrt{3}\)

\(=6+3\sqrt{15}-36\sqrt{5}\)

b) Ta có: \(\sqrt{252}-\sqrt{700}+\sqrt{1008}-\sqrt{448}\)

\(=3\sqrt{28}-5\sqrt{28}+3\sqrt{112}-2\sqrt{112}\)

\(=-2\sqrt{28}+\sqrt{112}=-\sqrt{112}+\sqrt{112}=0\)

c) Ta có: \(2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)

\(=2\cdot4\cdot\sqrt{5}\cdot\sqrt{\sqrt{3}}-2\cdot\sqrt{5}\cdot\sqrt{\sqrt{3}}-3\cdot2\cdot\sqrt{5}\cdot\sqrt{\sqrt{3}}\)

\(=8\sqrt{5}\cdot\sqrt{\sqrt{3}}-2\sqrt{5}\sqrt{\sqrt{3}}-6\sqrt{5}\sqrt{\sqrt{3}}\)

=0

Bài 2:

a) Ta có: \(A=\frac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}\)

\(=\frac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}\)

\(=\frac{1}{\sqrt{2}}\)

b) Ta có: \(B=\frac{9\sqrt{5}+3\sqrt{27}}{\sqrt{5}+\sqrt{3}}\)

\(=\frac{\sqrt{405}+\sqrt{243}}{\sqrt{5}+\sqrt{3}}\)

\(=\frac{9\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{5}+\sqrt{3}}=9\)

c) Ta có: \(C=\frac{3\sqrt{8}-2\sqrt{12}+\sqrt{20}}{3\sqrt{18}-2\sqrt{27}+\sqrt{45}}\)

\(=\frac{\sqrt{72}-\sqrt{48}+\sqrt{20}}{\sqrt{162}-\sqrt{108}+\sqrt{45}}\)

\(=\frac{2\left(\sqrt{18}-\sqrt{12}+\sqrt{5}\right)}{3\left(\sqrt{18}-\sqrt{12}+\sqrt{5}\right)}=\frac{2}{3}\)

a)

\(\sqrt{12}-\sqrt{27}+\sqrt{3}=\sqrt{4}.\sqrt{3}-\sqrt{9}.\sqrt{3}+\sqrt{3}=2\sqrt{3}-3\sqrt{3}+\sqrt{3}\)

\(=\sqrt{3}(2-3+1)=0\)

b)

\(\sqrt{252}-\sqrt{700}+\sqrt{1008}-\sqrt{448}=\sqrt{4}.\sqrt{63}-\sqrt{4}.\sqrt{175}+\sqrt{4}.\sqrt{252}-\sqrt{4}.\sqrt{112}\)

\(=2(\sqrt{63}-\sqrt{175}+\sqrt{252}-\sqrt{112})\)

\(=2(\sqrt{9}.\sqrt{7}-\sqrt{25}.\sqrt{7}+\sqrt{36}.\sqrt{7}-\sqrt{16}.\sqrt{7})\)

\(=2(3\sqrt{7}-5\sqrt{7}+6\sqrt{7}-4\sqrt{7})=2\sqrt{7}(3-5+6-4)=0\)

------------------

\(\sqrt{3}(\sqrt{12}+\sqrt{27}-\sqrt{3})=\sqrt{36}+\sqrt{81}-\sqrt{9}\)

\(=\sqrt{6^2}+\sqrt{9^2}-\sqrt{3^2}=6+9-3=12\)

c)

\(\frac{\sqrt{6}+\sqrt{10}}{\sqrt{21}+\sqrt{35}}=\frac{\sqrt{2}.\sqrt{3}+\sqrt{2}.\sqrt{5}}{\sqrt{7}.\sqrt{3}+\sqrt{7}.\sqrt{5}}=\frac{\sqrt{2}(\sqrt{3}+\sqrt{5})}{\sqrt{7}(\sqrt{3}+\sqrt{5})}=\frac{\sqrt{2}}{\sqrt{7}}\)

\(\frac{\sqrt{405}+3\sqrt{27}}{3\sqrt{3}+\sqrt{45}}=\frac{\sqrt{81}.\sqrt{5}+3\sqrt{9}.\sqrt{3}}{3\sqrt{3}+\sqrt{9}.\sqrt{5}}=\frac{9\sqrt{5}+9\sqrt{3}}{3\sqrt{3}+3\sqrt{5}}\)

\(=\frac{3(3\sqrt{5}+3\sqrt{3})}{3\sqrt{3}+3\sqrt{5}}=3\)

d)

\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}-\sqrt{6}-\sqrt{9}-\sqrt{12}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}-(\sqrt{6}+\sqrt{9}+\sqrt{12})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}-(\sqrt{2}.\sqrt{3}+\sqrt{3}.\sqrt{3}+\sqrt{3}.\sqrt{4})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}-\sqrt{3}(\sqrt{2}+\sqrt{3}+\sqrt{4})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{(\sqrt{2}+\sqrt{3}+\sqrt{4})(1-\sqrt{3})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1-\sqrt{3}\)

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