\(A=\left(\sqrt{22}+7\sqrt{2}\right)\sqrt{30-7\sqrt{11}}\)

\(2A=\left(\sqrt{44}+7\sqrt{4}\right)\sqrt{60-2.7\sqrt{11}}\)

\(2A=\left(2\sqrt{11}+14\right)\sqrt{7^2-2.7\sqrt{11}+\left(\sqrt{11}\right)^2}\)

\(2A=\left(2\sqrt{11}+14\right)\sqrt{\left(7-\sqrt{11}\right)^2}\)

\(2A=\left(2\sqrt{11}+14\right)\left|7-\sqrt{11}\right|\)

\(2A=\left(2\sqrt{11}+14\right)\left(7-\sqrt{11}\right)\)

\(A=\left(7+\sqrt{11}\right)\left(7-\sqrt{11}\right)\)

\(A=49-11=38\)

`a, (sqrt 28 - sqrt 12 - sqrt 7) sqrt 7 + 2 sqrt 21`.

`= sqrt(28.7) - sqrt(12.7) - sqrt(7.7) + 2 sqrt 21`.

`= sqrt(4. 7.7) - sqrt (12.7) - 7 + 2 sqrt 21`.

`= 14 - sqrt(4.3.7) - 7 + 2 sqrt 21`.

`= 7`.

`b, (sqrt99-sqrt18-sqrt11)sqrt11+3sqrt22`

`= sqrt(99.11)- sqrt(18.11)-sqrt(11.11) +3sqrt22`

`= sqrt(9.11.11)-sqrt(2.9.11)-11+3sqrt22`

`= 33 - 11 = 22`.

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

\(B=\sqrt{16+8\sqrt{7}+7}-\sqrt{7}\)

\(=\sqrt{\left(4+\sqrt{7}\right)^2}-\sqrt{7}\)

\(=4+\sqrt{7}-\sqrt{7}\\ =4\)

\(C=\sqrt{6+2\sqrt{30}+5}-\sqrt{6-2\sqrt{30}+5}\\ =\sqrt{\left(\sqrt{6}+\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}\\ =\sqrt{6}+\sqrt{5}-\sqrt{6}+\sqrt{5}\\ =2\sqrt{5}\)

a: \(=6-\sqrt{15}+2\sqrt{15}=6+\sqrt{15}\)

b: \(=\left(\sqrt{7}-2\sqrt{3}\right)\cdot\sqrt{7}+2\sqrt{21}\)

\(=7-2\sqrt{21}+2\sqrt{21}=7\)

c: \(=10+5\sqrt{10}-5\sqrt{10}=10\)

d: \(=22-\sqrt{198}+\sqrt{198}=22\)

a) \(\left(2\sqrt{3}+\sqrt{5}\right)\sqrt{3}-\sqrt{60}\) = \(6+\sqrt{15}-2\sqrt{15}\)

= \(6-\sqrt{15}\)

b) \(\left(5\sqrt{2}+2\sqrt{5}\right)\sqrt{5}-\sqrt{250}\) = \(5\sqrt{10}+10-5\sqrt{10}\) = \(10\)

c) \(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\) = \(14-2\sqrt{21}-7+2\sqrt{21}\)

= \(7\)

d) \(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)

= \(33-3\sqrt{22}-11+3\sqrt{22}\) = \(22\)

a)(2√3+√5)√3-√60
=6+√15-2√15
=6-√15

b)(5√2+2√5)√5-√250
=5√10+10-5√10
=10

c)(√28-√12-√7)√7+2√21
=14-2√21-7+2√21
=7

d)(√99-√18-√11)√11+3√22
=33-3√22-11+3√22
=22

\(\sqrt{7-\sqrt{24}}-\dfrac{\sqrt{50}-5}{\sqrt{10}-\sqrt{5}}+\sqrt{\left(11+\sqrt{120}\right)\left(11+2\sqrt{30}\right)^2}\)

\(=\sqrt{7-2\sqrt{6}}-\dfrac{5\left(\sqrt{2}-1\right)}{\sqrt{5}\left(\sqrt{2}-1\right)}+\left|11+2\sqrt{30}\right|\sqrt{11-2\sqrt{30}}\)

\(=\sqrt{1^2-2\sqrt{6}\cdot1+\left(\sqrt{6}\right)^2}-\dfrac{\sqrt{5}\cdot\sqrt{5}}{\sqrt{5}}+\left(11+2\sqrt{30}\right)\sqrt{\left(\sqrt{6}\right)^2-2\sqrt{5}\cdot\sqrt{6}+\left(\sqrt{5}\right)^2}\)

\(=\sqrt{\left(1-\sqrt{6}\right)^2}-\sqrt{5}+\left(11+2\sqrt{30}\right)\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}\)

\(=\left|1-\sqrt{6}\right|-\sqrt{5}+\left(11+2\sqrt{30}\right)\left|\sqrt{6}-\sqrt{5}\right|\)

\(=-1+6-\sqrt{5}+\left(\sqrt{6}+\sqrt{5}\right)^2\left(\sqrt{6}-\sqrt{5}\right)\)

\(=\sqrt{6}-1-\sqrt{5}+\left[\left(\sqrt{6}\right)^2-\left(\sqrt{5}\right)^2\right]\left(\sqrt{6}+\sqrt{5}\right)\)

\(=\sqrt{6}-1-\sqrt{5}+\left(6-5\right)\left(\sqrt{6}+\sqrt{5}\right)\)

\(=\sqrt{6}-1-\sqrt{5}+\sqrt{6}+\sqrt{5}\)

\(=2\sqrt{6}-1\)

\(=\sqrt{6+1-2\sqrt{6}}-\dfrac{\sqrt{5}\left(\sqrt{10}-\sqrt{5}\right)}{\sqrt{10}-\sqrt{5}}+\sqrt{\left(11-\sqrt{120}\right)\left(11+\sqrt{120}\right)^2}\\ =\sqrt{\left(\sqrt{6}-\sqrt{1}\right)^2}-\sqrt{5}+\sqrt{\left(11^2-120\right)\left(11+2\sqrt{30}\right)}\\ =\sqrt{6}-\sqrt{1}-\sqrt{5}+\sqrt{1\left(6+5+2\sqrt{6\cdot5}\right)}\\ =\sqrt{6}-\sqrt{1}-\sqrt{5}+\sqrt{\left(\sqrt{6}+\sqrt{5}\right)^2}\\ =\sqrt{6}-\sqrt{1}-\sqrt{5}+\sqrt{6}+\sqrt{5}=2\sqrt{6}-\sqrt{1}\)

Bn ghi đề vào nhé .