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TRẢ LỜI :
\(=\sqrt{5}+\sqrt{5}+\sqrt{5}=3\sqrt{5}\)
c) √20 - √45 + 3√18 + √72
= √4.5 - √9.5 + 3√9.2 + √36.2
= 2√5 - 3√5 + 9√2 + 6√2
= -√5 + 15√2
a: \(=3\sqrt{3}-2\sqrt{3}+4\sqrt{3}-5\sqrt{3}=2\sqrt{3}\)
a: \(=3\sqrt{5}+2\sqrt{5}-2\sqrt{5}=3\sqrt{5}\)
b: \(=2\sqrt{2}+2\sqrt{2}+5\sqrt{2}=9\sqrt{2}\)
c: \(=4\sqrt{3}+3\sqrt{3}-3\sqrt{5}+2\sqrt{5}=7\sqrt{3}-\sqrt{5}\)
d: \(=5\sqrt{3}+4\sqrt{3}-10\sqrt{3}=-\sqrt{3}\)
e: \(=\left(\sqrt{7}-2\sqrt{3}\right)\cdot\sqrt{7}+2\sqrt{21}\)
=7-2*căn 21+2*căn 21
=7
f: \(=\left(2\sqrt{11}-3\sqrt{2}\right)\cdot\sqrt{11}+3\sqrt{22}\)
=22-3*căn 22+3*căn 22
=22
a) \(3\sqrt{5}+\sqrt{20}-2\sqrt{5}\)
\(=3\sqrt{5}+2\sqrt{5}-2\sqrt{5}\)
\(=3\sqrt{5}\)
b) \(2\sqrt{2}+\sqrt{8}+\sqrt{50}\)
\(=2\sqrt{2}+2\sqrt{2}+5\sqrt{2}\)
\(=9\sqrt{5}\)
c) \(4\sqrt{3}+\sqrt{27}-\sqrt{45}+2\sqrt{5}\)
\(=4\sqrt{3}+3\sqrt{3}-3\sqrt{5}+2\sqrt{5}\)
\(=7\sqrt{3}-\sqrt{5}\)
d) \(\sqrt{75}+\sqrt{48}-\sqrt{300}\)
\(=5\sqrt{3}+4\sqrt{3}-10\sqrt{3}\)
\(=-\sqrt{3}\)
e) \(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\)
\(=\left(2\sqrt{7}-2\sqrt{3}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\)
\(=\left(\sqrt{7}-2\sqrt{3}\right)\sqrt{7}+2\sqrt{21}\)
\(=7-2\sqrt{21}+2\sqrt{21}\)
\(=7\)
f) \(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)
\(=\left(3\sqrt{11}-3\sqrt{2}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)
\(=\left(2\sqrt{11}-3\sqrt{2}\right)\sqrt{11}+3\sqrt{22}\)
\(=22-3\sqrt{22}+3\sqrt{22}\)
\(=22\)
g) \(3\sqrt{45}-5\sqrt{125x}+7\sqrt{20x}+28\)
\(=9\sqrt{5}-25\sqrt{5x}+14\sqrt{5x}+28\)
\(=9\sqrt{5}-11\sqrt{5x}+28\)
1.
A= \(2\sqrt{6}\) + \(6\sqrt{6}\) - \(8\sqrt{6}\)
A= 0
2.
A= \(12\sqrt{3}\) + \(5\sqrt{3}\) - \(12\sqrt{3}\)
A= 0
3.
A= \(3\sqrt{2}\) - \(10\sqrt{2}\) + \(6\sqrt{2}\)
A= -\(\sqrt{2}\)
4.
A= \(3\sqrt{2}\) + \(4\sqrt{2}\) - \(\sqrt{2}\)
A= \(6\sqrt{2}\)
5.
M= \(2\sqrt{5}\) - \(3\sqrt{5}\) + \(\sqrt{5}\)
M= 0
6.
A= 5 - \(3\sqrt{5}\) + \(3\sqrt{5}\)
A= 5
This literally took me a while, pls sub :D
https://www.youtube.com/channel/UC4U1nfBvbS9y_Uu0UjsAyqA/featured
a) \(\dfrac{1}{2}\sqrt{20}+5=\dfrac{1}{2}\cdot2\sqrt{5}+5=5+\sqrt{5}\)
b) \(\sqrt{16}+\sqrt{64}=4+8=12\)
c) \(\sqrt{20}-\sqrt{45}+3\sqrt{18}=2\sqrt{5}-3\sqrt{5}+9\sqrt{2}=9\sqrt{2}-\sqrt{5}\)
d) \(\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{2}=2-\sqrt{2}+\sqrt{2}=2\)
\(a,=3\sqrt{2}-12\sqrt{2}+8\sqrt{2}-5\sqrt{2}\)
\(=\sqrt{2}\left(3-12+8-5\right)=-6\sqrt{2}\)
\(b,=\left|\sqrt{2}-\sqrt{3}\right|+3\sqrt{2}=\sqrt{3}-\sqrt{2}+3\sqrt{2}=\sqrt{3}+2\sqrt{2}\)
\(c,=\sqrt{5}+\sqrt{5}+\dfrac{5}{\sqrt{5}}-1=3\sqrt{5}-1\)
\(d,=\sqrt{3-2.2\sqrt{3}+4}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+1+\sqrt{3}=2\)
a) \(3\sqrt{2}-4\sqrt{18}+2\sqrt{32}-\sqrt{50}=3\sqrt{2}-4\sqrt{9.2}+2\sqrt{16.2}-\sqrt{25.2}\)
\(=3\sqrt{2}-12\sqrt{2}+8\sqrt{2}-5\sqrt{2}=-6\sqrt{2}\)
b) \(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}+\sqrt{18}=\left|\sqrt{2}-\sqrt{3}\right|+\sqrt{9.2}=\sqrt{3}-\sqrt{2}+3\sqrt{2}\)
\(=2\sqrt{2}+\sqrt{3}\)
c) \(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{45}+\dfrac{5-\sqrt{5}}{\sqrt{5}}=\sqrt{25.\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{9.5}+\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}}\)
\(=\sqrt{5}+\sqrt{5}+\sqrt{5}-1=3\sqrt{5}-1\)
d) \(\sqrt{7-4\sqrt{3}}+\sqrt{\left(1+\sqrt{3}\right)^2}=\sqrt{2^2-2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}+\left|\sqrt{3}+1\right|\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{3}+1=\left|2-\sqrt{3}\right|+\sqrt{3}+1=2-\sqrt{3}+\sqrt{3}+1=3\)
= (2√2 - 3√2 + 10)√2 - √5
= 2.(√2)2 - 3.(√2)2 + √10.√2 - √5
= 4 - 6 + √20 - √5 = -2 + 2√5 - √5
= -2 + √5
= 0,2.10.√3 + 2|√3 - √5|
s
= 2√3 + 2(√5 - √3)
= 2√3 + 2√5 - 2√3 = 2√5
= √5 + √5 + √5 = 3√5
c) √20 - √45 + 3√18 + √72
= √4.5 - √9.5 + 3√9.2 + √36.2
= 2√5 - 3√5 + 9√2 + 6√2
= -√5 + 15√2