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\(\left(2+\sqrt{3}\right)\left(\sqrt{7-4\sqrt{3}}\right)=\left(2+\sqrt{3}\right)\sqrt{4-4\sqrt{3}+3}\)
\(=\left(2+\sqrt{3}\right).\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\left(2+\sqrt{3}\right)\left|2-\sqrt{3}\right|\)
\(=\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)\)( Vì \(2-\sqrt{3}>0\))
\(=4-2=1\)
mk
\(\sqrt{\frac{4}{9-4\sqrt{5}}}+\sqrt{\frac{9}{9+4\sqrt{5}}}=\sqrt{\frac{4}{4-4\sqrt{5}+5}}+\sqrt{\frac{9}{4+4\sqrt{5}+5}}\)
\(=\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}+\sqrt{\frac{9}{\left(2+\sqrt{5}\right)^2}}=\frac{2}{\sqrt{5}-2}+\frac{3}{2+\sqrt{5}}\)
\(=\frac{2\left(\sqrt{5}+2\right)}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}+\frac{3\left(2-\sqrt{5}\right)}{\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)}=\frac{2\sqrt{5}+4}{5-4}+\frac{6-3\sqrt{5}}{4-5}\)
\(=2\sqrt{5}+4+3\sqrt{5}-6=5\sqrt{5}-2\)
b) \(\left(5-4\sqrt{3}\right):\frac{2+\sqrt{3}}{2-\sqrt{3}}=\left(5-4\sqrt{3}\right).\frac{2-\sqrt{3}}{2+\sqrt{3}}\)
\(=\left(5-4\sqrt{3}\right).\frac{\left(2-\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\left(5-4\sqrt{3}\right).\frac{4-4\sqrt{3}+3}{4-3}\)
\(=\left(5-4\sqrt{3}\right)\left(7-4\sqrt{3}\right)=35-28\sqrt{3}-20\sqrt{3}+48\)
\(=73-48\sqrt{3}\)
Mình chịu câu c nha
a: \(=2\sqrt{6}-4\sqrt{2}+9+4\sqrt{2}-2\sqrt{6}=9\)
b: \(=\sqrt{81-17}=8\)
trinh mai
\(\sqrt{\left(\sqrt{2}-3\right)^2}.\sqrt{3^2+3.2\sqrt{2}+2}=\sqrt{\left(3-\sqrt{2}\right)^2}.\sqrt{\left(3+\sqrt{2}\right)^2}=\left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right)=3^2-2=7\)
\(a,\sqrt{17-4\sqrt{9+4\sqrt{5}}}=\sqrt{17-4\sqrt{5+4\sqrt{5}+4}}=\sqrt{17-4\sqrt{\left(\sqrt{5}\right)^2+2.2\sqrt{5}+2^2}}=\sqrt{17-4\sqrt{\sqrt{\left(\sqrt{5}+2\right)^2}}}=\sqrt{17-4\sqrt{\sqrt{5}+2}}\) \(b,\sqrt{a};đk:a\ge0;2-3=-1< 0\Rightarrow sai\)
\(c,\sqrt{\left(\sqrt{3-3}\right)^2}.\sqrt{\frac{1}{3-\sqrt{3}}}=\sqrt{0^2}.\sqrt{\frac{1}{3-\sqrt{3}}}=0.\sqrt{\frac{1}{3-\sqrt{3}}}=0\)
\(d,\left(\sqrt{6}-3\sqrt{3}+5\sqrt{2}-\frac{1}{2}\sqrt{8}\right)2\sqrt{6}=\left(\sqrt{2}.\sqrt{3}-3\sqrt{3}+5\sqrt{2}-\sqrt{2}\right)2\sqrt{6}=\left[\sqrt{3}\left(\sqrt{2}-3\right)+\sqrt{2}.4\right]2\sqrt{6}=\left[2.\sqrt{3}.\sqrt{2}.\sqrt{3}\left(\sqrt{2}-3\right)+\sqrt{2}.\sqrt{2}.\sqrt{3}.2.4\right]=6\sqrt{2}\left(\sqrt{2}-3\right)+16\sqrt{3}\)
Bài 1:
a: \(=\sqrt{32.4}=\dfrac{9}{5}\sqrt{10}\)
b: \(=\sqrt{5\cdot5\cdot7\cdot7\cdot11\cdot11}=5\cdot7\cdot11=385\)
c: \(=5-2\sqrt{6}\)
d: \(=18-1=17\)
e: \(=3\sqrt{2}-2\sqrt{3}+7\sqrt{3}-7\sqrt{2}=-4\sqrt{2}+5\sqrt{3}\)
a) \(\sqrt{3^2}-\sqrt{7^2}+\sqrt{\left(-1\right)^2}=|3|-|7|+|-1|=3-7+1=-3\)
b) \(-2\sqrt{\left(-2\right)^2}+\sqrt{\left(-5\right)^2}+\sqrt{3^2}=-2|2|+|-5|+\left|3\right|=-4+5+3=4\)
c) \(\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(2+\sqrt{2}\right)^2}=\left|2-\sqrt{2}\right|+\left|2+\sqrt{2}\right|=2-\sqrt{2}+2+\sqrt{2}=4\)
d) \(\sqrt{\left(3\sqrt{2}\right)^2}-\sqrt{\left(1-\sqrt{2}\right)^2}=\left|3\sqrt{2}\right|-\left|1-\sqrt{2}\right|=3\sqrt{2}-\sqrt{2}+1=2\sqrt{2}+1\)
e) \(\sqrt{3-2\sqrt{2}}+\sqrt{3+2\sqrt{2}}=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(\sqrt{2}+1\right)^2}=\left|\sqrt{2}-1\right|+\left|\sqrt{2}+1\right|=\sqrt{2}-1+\sqrt{2}+1=2\sqrt{2}\)
f) \(\sqrt{9-4\sqrt{5}}+\sqrt{9+4\sqrt{5}}=\sqrt{\left(\sqrt{5}-2\right)^2}+\sqrt{\left(\sqrt{5}+2\right)^2}=\left|\sqrt{5}-2\right|+\left|\sqrt{5}+2\right|=\sqrt{5}-2+\sqrt{5}+2=2\sqrt{5}\)
g) \(\sqrt{9-4\sqrt{2}}+\sqrt{11-6\sqrt{2}}=\sqrt{9-2\sqrt{8}}+\sqrt{2-2\sqrt{2}.3+9}=\sqrt{\left(\sqrt{8}-1\right)^2}+\sqrt{\left(\sqrt{2}-3\right)^2}=\sqrt{8}-1+3-\sqrt{2}=2-\sqrt{2}+\sqrt{8}\)
h) \(\sqrt{12+8\sqrt{2}}+\sqrt{6-4\sqrt{2}}=\sqrt{12+2\sqrt{4}\sqrt{8}}+\sqrt{6-2\sqrt{2}\sqrt{4}}=\sqrt{\left(\sqrt{4}+\sqrt{8}\right)^2}+\sqrt{\left(\sqrt{4}-\sqrt{2}\right)^2}=\sqrt{4}+\sqrt{8}+\sqrt{4}-\sqrt{2}\)
k) \(\left(2-\sqrt{3}\right)\sqrt{7+4\sqrt{3}}=\left(2-\sqrt{3}\right)\sqrt{\left(\sqrt{3}+2\right)^2}=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=4-3=1\)
Không hiểu sao cứ gửi ảnh nó lại bị lộn xộn nên bạn cố nhìn nhé
( ͡°( ͡° ͜ʖ( ͡° ͜ʖ ͡°)ʖ ͡°) ͡°)