`a, (2 sqrt 3 + sqrt 5)sqrt 3 - sqrt 60`

`= 2 sqrt 3 . sqrt 3 + sqrt 5 . sqrt 3 - sqrt(4 . 15)`

`= 2 . 3 + sqrt 15 - 2 sqrt 15`.

`= 6 - sqrt 15`.

`b, (5 sqrt 2 + 2 sqrt 5)sqrt 5 - sqrt250`

`= 5 sqrt 2 . sqrt 5 + 2 sqrt 5 . sqrt 5 - sqrt(25.10)`

`= 5 sqrt 10 + 10 - 5 sqrt 10`

`= 10`.

5) \(\left(2\sqrt{3}+\sqrt{5}\right)\sqrt{3}-\sqrt{60}\)

\(=2\sqrt{3}\cdot\sqrt{3}+\sqrt{5}\cdot\sqrt{3}-\sqrt{2^2\cdot15}\)

\(=2\cdot3+\sqrt{15}-2\sqrt{15}\)

\(=6+\left(1-2\right)\sqrt{15}\)

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

6) \(\left(5\sqrt{2}+2\sqrt{5}\right)\sqrt{5}-\sqrt{250}\)

\(=5\sqrt{2}\cdot\sqrt{5}+2\sqrt{5}\cdot\sqrt{5}-\sqrt{5^2\cdot10}\)

\(=5\sqrt{10}+2\cdot5-5\sqrt{10}\)

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

\(=0+10\)

\(=10\)

a) \(2\sqrt{98}-3\sqrt{18}+\dfrac{1}{2}\sqrt{32}=14\sqrt{2}-9\sqrt{2}+2\sqrt{2}=7\sqrt{2}\)

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

c) \(\left(2\sqrt{3}-5\sqrt{2}\right).\sqrt{3}-\sqrt{36}=6-5\sqrt{6}-6=5\sqrt{6}\)

d) \(3\sqrt{48}+2\sqrt{27}-\dfrac{1}{3}\sqrt{243}=12\sqrt{3}+6\sqrt{3}-3\sqrt{3}=15\sqrt{3}\)

e) \(6\sqrt{\dfrac{1}{3}}+\dfrac{9}{\sqrt{3}}-\dfrac{2}{\sqrt{3}-1}=2\sqrt{3}+3\sqrt{3}=\left(\sqrt{3}+1\right)=4\sqrt{3}-1\)

f) \(4\sqrt{\dfrac{1}{2}}-\dfrac{6}{\sqrt{2}}.\dfrac{2}{\sqrt{2}+1}=2\sqrt{2}-\left(12-6\sqrt{2}\right)=8\sqrt{2}-12\)

\(\text{a)}\)\(\left(5\sqrt{2}+2\sqrt{5}\right)\sqrt{5}-\sqrt{250}\)

\(\Leftrightarrow5\sqrt{10}+10-\sqrt{250}\)

\(\Leftrightarrow5\sqrt{10}+10-5\sqrt{10}\)

\(\Leftrightarrow10\)

\(\text{b)}\)\(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\)

\(\Leftrightarrow4\sqrt{21}-2\sqrt{21}-7+2\sqrt{21}\)

\(\Leftrightarrow4\sqrt{21}-7\)

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

a)

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

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

\(=\sqrt{5}.2\sqrt{5}=10\)

b)

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

\(=5+2\sqrt{6}+5-2\sqrt{6}=10\)

c)

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

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

d)

\(\dfrac{\sqrt{x-1-2\sqrt{x-2}}}{\sqrt{x-2}-1}\)

\(=\dfrac{\left(\sqrt{x-2}-1\right)^2}{\sqrt{x-2}-1}=\sqrt{x-2}-1\)

\(\left(5\sqrt{2}+2\sqrt{5}\right)\sqrt{5}-\sqrt{250}\)
\(=5\sqrt{10}+2.5-\sqrt{\left(25.10\right)}\)
\(=5\sqrt{10}+10-5\sqrt{10}\)
\(=10\)

Giải:

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

\(=5\sqrt{10}+10-\sqrt{250}\)

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

\(=10\)

Vậy ...

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: \(=22-\sqrt{198}+\sqrt{198}=22\)

1) \(\left(5\sqrt{2}+2\sqrt{5}\right)\sqrt{5}-\sqrt{250}\)

\(=5\sqrt{10}-10-5\sqrt{10}\)

\(=-10\)

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

\(=14-2\sqrt{21}-7+2\sqrt{21}\)

\(=7\)

3) \(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\) (hẳn đề là như thế này)

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

\(=22\)