a, = \(3\sqrt{2}+3\sqrt{3}+9\sqrt{2}-5\sqrt{3}\)

= \(12\sqrt{2}-2\sqrt{3}\)

b, = 5 - 7 - 8 + 2

= - 8

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) \(=\sqrt{\frac{9}{2}}-\sqrt{16.2}+\sqrt{36.2}-\sqrt{81.2}\)

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

\(=\left(\frac{3}{2}-4+6-9\right)\sqrt{2}=\frac{-11}{2}\sqrt{2}\)

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

\(=\frac{6}{5-9}.\left(-\sqrt{3}\right)=\frac{3}{2}\sqrt{3}\)

c) \(=\left(\frac{a-1-4\sqrt{a}+\sqrt{a}+1}{a-1}\right):\frac{\sqrt{a}\left(\sqrt{a}-2\right)}{a-1}\)

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

\(=\frac{\sqrt{a}\left(\sqrt{a}-3\right)}{\sqrt{a}\left(\sqrt{a}-2\right)}=\frac{\sqrt{a}-3}{\sqrt{a}-2}\)

\(A=6\sqrt{27}-2\sqrt{75}-\frac{1}{2}\sqrt{300}\)

\(A=6\sqrt{3^2.3}-2\sqrt{5^2.3}-\frac{1}{2}\sqrt{10^2.3}\)

\(A=18\sqrt{3}-10\sqrt{3}-5\sqrt{3}\)

\(A=3\sqrt{3}\)

vậy \(A=3\sqrt{3}\)

\(B=\left(1+\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1-\frac{x-\sqrt{x}}{\sqrt{x}-1}\right)\)  \(ĐKXĐ:x>0;x\ne1\)

\(B=\left[1+\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right]\left[1+\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\)

\(B=\left[1+\sqrt{x}\right]\left[1-\sqrt{x}\right]\)

\(B=1-x\)

vậy \(B=1-x\)

\(C=\sqrt[3]{64}-\sqrt[3]{-125}+\sqrt[3]{216}\)

\(C=\sqrt[3]{4^3}-\sqrt[3]{\left(-5\right)^3}+\sqrt[3]{6^3}\)

\(C=4+5+6\)

\(C=15\)

vậy \(C=15\)

Cho mk giải câu a:

\(A=6\sqrt{27}-2\sqrt{75}-\frac{1}{2}\sqrt{300}\)

\(A=18\sqrt{3}-10\sqrt{3}-\frac{1}{2}10\sqrt{3}\)

\(A=18\sqrt{3}-10\sqrt{3}-10:2\sqrt{3}\)

\(A=18\sqrt{3}-10\sqrt{3}-5\sqrt{3}\)

\(A=\left(18-10-5\right)\sqrt{3}\)

\(A=3\sqrt{3}\)

a) \(\sqrt{200}+2\sqrt{108}-\sqrt{98}+\frac{1}{3}\sqrt{\frac{81}{3}}-3\sqrt{75}\)

\(=10\sqrt{2}+12\sqrt{3}-7\sqrt{2}+\sqrt{3}-15\sqrt{3}\)

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

b)\(\left(21\sqrt{\frac{1}{7}}+\frac{1}{2}\sqrt{112}-\frac{14}{3}\sqrt{\frac{9}{7}}+7\right):3\sqrt{7}\)

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

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

c)\(\left(\sqrt{27}-\sqrt{125}+\sqrt{45}+\sqrt{12}\right)\left(\sqrt{75}+\sqrt{20}\right)\)

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

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

\(=75-20=55\)

d)\(\left(\frac{3}{\sqrt{6}-3}-\frac{3}{\sqrt{6}+3}\right).\frac{3-\sqrt{3}}{2-2\sqrt{3}}-\frac{\sqrt{28-6\sqrt{3}}}{1}\)

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

\(=\frac{-6\left(3-\sqrt{3}\right)}{2-2\sqrt{3}}-\left(3\sqrt{3}-1\right)\left(do3\sqrt{3}>1\right)\)

\(=\frac{6\sqrt{3}-18}{2-2\sqrt{3}}-\frac{8\sqrt{3}-20}{2-2\sqrt{3}}\)

\(=\frac{6\sqrt{3}-18-8\sqrt{3}+20}{2-2\sqrt{3}}=\frac{2-2\sqrt{3}}{2-2\sqrt{3}}=1\)