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a) \(\sqrt{20}+2\sqrt{45}-3\sqrt{80}+\sqrt{125}=\sqrt{4.5}+2\sqrt{9.5}-3\sqrt{16.5}+\sqrt{25.5}=2\sqrt{5}+6\sqrt{5}-12\sqrt{6}+5\sqrt{5}=\sqrt{5}\) b) \(\sqrt{6+2\sqrt{5}}-\sqrt{21+4\sqrt{5}}+\sqrt{5}\left(\sqrt{5}+1\right)=\sqrt{5+2\sqrt{5}+1}-\sqrt{20+2.2\sqrt{5}+1}+\sqrt{5}\left(\sqrt{5}+1\right)=\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(2\sqrt{5}+1\right)^2}+\sqrt{5}\left(\sqrt{5}+1\right)\) = / \(\sqrt{5}+1\) / + / \(2\sqrt{5}+1\) / \(+\sqrt{5}\left(\sqrt{5}+1\right)\)
\(=\sqrt{5}+1\) + \(2\sqrt{5}+1\) \(+\sqrt{5}\left(\sqrt{5}+1\right)\)
= \(4\sqrt{5}+7\)
Bạn mới trả lời câu a,b thôi đúng không?Nhưng mà rất cảm ơn bạn!:)))
a) \(\left(3+1\sqrt{6}-\sqrt{33}\right)\left(\sqrt{22}+\sqrt{6}+4\right)\)
\(=\sqrt{3}\left(\sqrt{3}+2\sqrt{2}-\sqrt{11}\right).\sqrt{2}\left(\sqrt{11}+\sqrt{3}+2\sqrt{2}\right)\)
\(=\sqrt{6}\left(\sqrt{3}+2\sqrt{2}-\sqrt{11}\right)\left(\sqrt{3}+2\sqrt{2}+\sqrt{11}\right)\)
\(=\sqrt{6}\left[\left(\sqrt{3}+2\sqrt{2}\right)^2-11\right]=\sqrt{6}\left(11+4\sqrt{6}-11\right)=\sqrt{6}.4\sqrt{6}=6.4=24\)
b) \(\left(\frac{1}{5-2\sqrt{6}}+\frac{2}{5+2\sqrt{6}}\right)\left(15+2\sqrt{6}\right)=\left(\frac{5+2\sqrt{6}+10-4\sqrt{6}}{5^2-\left(2\sqrt{6}\right)^2}\right)\left(15+2\sqrt{6}\right)\)
\(=\left(15-2\sqrt{6}\right)\left(15+2\sqrt{6}\right)=15^2-24=201\)
C) \(\left(\frac{4}{3}.\sqrt{3}+\sqrt{2}+\sqrt{3\frac{1}{3}}\right)\left(\sqrt{1,2}+\sqrt{2}-4\sqrt{\frac{1}{5}}\right)\)
\(=\left(\frac{4}{\sqrt{3}}+\frac{\sqrt{6}}{\sqrt{3}}+\frac{\sqrt{10}}{\sqrt{3}}\right)\left(\frac{\sqrt{6}}{\sqrt{5}}+\frac{\sqrt{10}}{\sqrt{5}}-\frac{4}{\sqrt{5}}\right)\)
\(=\frac{1}{\sqrt{15}}\left(\sqrt{6}+\sqrt{10}+4\right)\left(\sqrt{6}+\sqrt{10}-4\right)=\frac{1}{\sqrt{15}}\left[\left(\sqrt{6}+\sqrt{10}\right)^2-16\right]\)
\(=\frac{1}{\sqrt{15}}\left(16+4\sqrt{15}-16\right)=\frac{4\sqrt{15}}{\sqrt{15}}=4\)
d) \(\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{1990+2\sqrt{1989}}=\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{1989+2\sqrt{1989}+1}\)
\(=\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{\left(\sqrt{1989}+1\right)^2}=\left(\sqrt{1989}-1\right)\left(\sqrt{1989}+1\right)=1989-1=1988\)
e) \(\frac{a-\sqrt{ab}+b}{a\sqrt{a}+b\sqrt{b}}-\frac{1}{a-b}=\frac{a-\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}-\frac{1}{a-b}=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}-\frac{1}{a-b}=\frac{\sqrt{a}-\sqrt{b}-1}{a-b}\)
a) Ta có: \(3\sqrt{2}+4\sqrt{8}-\sqrt{18}\)
\(=\sqrt{2}\left(3+4\cdot2-3\right)\)
\(=8\sqrt{2}\)
b) Ta có: \(\sqrt{3}-\frac{1}{3}\sqrt{27}+2\sqrt{507}\)
\(=\sqrt{3}\left(1-\frac{1}{3}\cdot\sqrt{9}+2\cdot\sqrt{169}\right)\)
\(=\sqrt{3}\left(1-1+26\right)\)
\(=26\sqrt{3}\)
c) Ta có: \(\sqrt{25a}+\sqrt{49a}-\sqrt{64a}\)
\(=\sqrt{25}\cdot\sqrt{a}+\sqrt{49}\cdot\sqrt{a}-\sqrt{64}\cdot\sqrt{a}\)
\(=\sqrt{a}\left(5+7-8\right)\)
\(=4\sqrt{a}\)
d) Ta có: \(-\sqrt{36b}-\frac{1}{3}\sqrt{54b}+\frac{1}{5}\sqrt{150b}\)
\(=-\sqrt{6b}\cdot\sqrt{6}-\frac{1}{3}\cdot\sqrt{6b}\cdot\sqrt{9}+\frac{1}{5}\cdot\sqrt{6b}\cdot\sqrt{25}\)
\(=-\sqrt{6b}\left(\sqrt{6}+1-1\right)\)
\(=-\sqrt{6b}\cdot\sqrt{6}=-6\sqrt{b}\)
a) \(\sqrt{17}-4\) b) \(\sqrt{3}\) c) \(\frac{\sqrt{2}}{2}\) d)\(\frac{\sqrt{x}+1}{\sqrt{x}-1}\) e) \(x-\sqrt{5}\)
f) \(4+2\sqrt{3}\) g) \(3+2\sqrt{2}\) h) \(x+\sqrt{x}+1\) i) \(\frac{3\sqrt{5}-\sqrt{15}}{10}\)
k) \(\sqrt{5}+\sqrt{6}\) i) 5 h) 0 l) \(\sqrt{5}+\sqrt{3}\) m) \(\frac{20\sqrt{3}}{3}\) d) 0
1, \(\sqrt{8}-3\sqrt{32}+\sqrt{72}=2\sqrt{2}-12\sqrt{2}+6\sqrt{2}=-4\sqrt{2}\)
2,\(6\sqrt{12}-2\sqrt{48}+5\sqrt{75}-7\sqrt{108}=12\sqrt{3}-8\sqrt{3}+25\sqrt{3}-42\sqrt{3}=-13\sqrt{3}\)
3, \(\sqrt{20}+3\sqrt{45}-6\sqrt{80}-\dfrac{1}{3}\sqrt{125}=2\sqrt{5}+9\sqrt{5}-24\sqrt{5}-\dfrac{5}{3}.\sqrt{5}=-\dfrac{44}{3}.\sqrt{5}\)
4, \(2\sqrt{5}-\sqrt{125}-\sqrt{80}=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}=-7\sqrt{5}\)
5, \(3\sqrt{2}-\sqrt{8}+\sqrt{50}-4\sqrt{32}=3\sqrt{2}-2\sqrt{2}+5\sqrt{2}-16\sqrt{2}=-10\sqrt{2}\)
a) \(\sqrt{49.360}=\sqrt{7^2.6^2.10}=7.6\sqrt{10}=42\sqrt{10}\)
b)\(\sqrt{125a^2}=\sqrt{5^2.5.a^2}=5.\left|a\right|\sqrt{5}=-5a\sqrt{5}\) ( vì a<0)
c)\(-\sqrt{500.162}=-\sqrt{10^2.5.9^2.2}=-10.9\sqrt{5.2}=-90\sqrt{10}\)
d) \(\frac{1}{3}\sqrt{225a^2}=\frac{1}{3}\sqrt{15^2.a^2}=\frac{1}{3}.15.\left|a\right|=\frac{15a}{3}\) ( a>0)
a: \(=-6\sqrt{b}-\dfrac{1}{3}\cdot3\sqrt{3b}+\dfrac{1}{5}\cdot5\sqrt{6b}\)
\(=-6\sqrt{b}-\sqrt{3}\cdot\sqrt{b}+\sqrt{6}\cdot\sqrt{b}\)
\(=\sqrt{b}\left(-6-\sqrt{3}+\sqrt{6}\right)\)
c: \(=\sqrt{\left(5+2\sqrt{6}\right)^2}+\sqrt{\left(5-2\sqrt{6}\right)^2}\)
\(=5+2\sqrt{6}+5-2\sqrt{6}=10\)
d: \(A=\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)
e: \(B=\sqrt{6+2\sqrt{5-2\sqrt{3}-1}}\)
\(=\sqrt{6+2\cdot\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)
1. \(2\sqrt{5}-5\sqrt{20}+\sqrt{80}\)
= \(2\sqrt{5}-5.2\sqrt{5}+4\sqrt{5}\)
= \(2\sqrt{5}-10\sqrt{5}+4\sqrt{5}\)
= \(-4\sqrt{5}\)
2. B = \(\frac{1}{\sqrt{5}-2}-\sqrt{6-2\sqrt{5}}\)
= \(\frac{1}{\sqrt{5}-2}-\sqrt{1^2-2\sqrt{5}+\sqrt{15}^2}\)
= \(\frac{1}{\sqrt{5}-2}-\sqrt{\left(1-\sqrt{5}\right)^2}\)
= \(\frac{1}{\sqrt{5}-2}-\left|1-\sqrt{5}\right|\)
= \(\frac{1}{\sqrt{5}-2}-\sqrt{5}+1\left(\sqrt{5}>1\right)\)
= \(\frac{1}{\sqrt{5}-2}-\frac{\left(\sqrt{5}+1\right)\left(\sqrt{5}-2\right)}{\sqrt{5}-2}=\frac{1-5+2\sqrt{5}+\sqrt{5}-2}{\sqrt{5}-2}\)
= \(\frac{-6+3\sqrt{5}}{\sqrt{5}-2}=\frac{3\left(\sqrt{5}-2\right)}{\sqrt{5}-2}=3\)