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B1 :
a) \(\sqrt{1,2.270}=\sqrt{0,4.3.90.3}=3\sqrt{36}=3.6=18\)
\(\sqrt{55.77.35}=\sqrt{5.11.7.11.7.5}=\sqrt{25.49.212}=\sqrt{25}.\sqrt{49}.\sqrt{121}=5.7.11=385\)
b) \(\left(\sqrt{3}-\sqrt{2}\right)^2=3-2.\sqrt{3}.\sqrt{2}+2=5-2\sqrt{6}\)
\(\left(3\sqrt{2}-1\right)\left(3\sqrt{2}+1\right)=3\sqrt{2}.3\sqrt{2}+3\sqrt{2}-3\sqrt{2}-1=18-1\)
\(\left(\sqrt{6}+2\right)\left(\sqrt{3}-2\right)=\sqrt{6}.\sqrt{3}-2\sqrt{6}+2\sqrt{3}-4=\sqrt{18}-2\sqrt{6}+2\sqrt{3}-4\)\(=3\sqrt{2}-2\sqrt{6}+2\sqrt{3}-4\)
\(c,\left(\sqrt{\dfrac{3}{2}}-\sqrt{\dfrac{2}{3}}\right)=\dfrac{\sqrt{3}}{\sqrt{2}}-\dfrac{\sqrt{2}}{\sqrt{3}}=\dfrac{3-2}{\sqrt{2}\sqrt{3}}\) = \(\dfrac{1}{\sqrt{6}}\)
\(\left(\sqrt{\dfrac{8}{3}}-\sqrt{24}+\sqrt{\dfrac{50}{3}}\right).\sqrt{6}=\sqrt{\dfrac{8}{3}}.\sqrt{6}-\sqrt{24}.\sqrt{6}+\sqrt{\dfrac{50}{3}}.\sqrt{6}\) = \(\dfrac{\sqrt{8}.\sqrt{6}}{\sqrt{3}}-\sqrt{144}+\dfrac{\sqrt{50}.\sqrt{6}}{\sqrt{3}}=\dfrac{\sqrt{48}}{\sqrt{3}}-12+\dfrac{\sqrt{300}}{\sqrt{3}}=\sqrt{\dfrac{48}{3}}-12+\sqrt{\dfrac{300}{3}}=4-12+10=2\)
B2 :
a) \(\sqrt{\dfrac{1}{8}}.\sqrt{2}.\sqrt{125}.\sqrt{\dfrac{1}{5}}=\sqrt{\dfrac{1}{8}.2.125.\dfrac{1}{5}}=\sqrt{\dfrac{25}{4}}=\dfrac{5}{2}\)
\(\sqrt{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\sqrt{2+\sqrt{2}-\sqrt{2}-1}=1\)
b) \(\sqrt{\left(\sqrt{2}-3\right)^2}.\sqrt{11+6\sqrt{2}}=\left|\sqrt{2}-3\right|.\sqrt{2+6\sqrt{2}+9}=\left(\sqrt{2}-3\right).\sqrt{\left(\sqrt{2}+3\right)^2}=\left(\sqrt{2}-3\right)\)\(\left(\sqrt{2}+3\right)=2+3\sqrt{2}-3\sqrt{2}-9=-7\)
\(\sqrt{\left(\sqrt{3}-3\right)^2}.\sqrt{\dfrac{1}{3-\sqrt{3}}}=\left|\sqrt{3}-3\right|.\dfrac{1}{3-\sqrt{3}}=-\left(3-\sqrt{3}\right).\left(\dfrac{1}{3-\sqrt{3}}\right)=-1\)
\(\left(3\sqrt{2}+\sqrt{6}\right)\left(6-3\sqrt{3}\right)\)
\(=\sqrt{6}\left(\sqrt{3}+1\right)\times3\left(2-\sqrt{3}\right)\)
\(=\dfrac{3\sqrt{6}}{2}\left(\sqrt{3}+1\right)\left(4-2\sqrt{3}\right)\)
\(=\dfrac{3\sqrt{6}}{2}\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)^2\)
\(=\dfrac{3\sqrt{6}}{2}\left(3-1\right)\left(\sqrt{3}-1\right)\)
\(=3\sqrt{6}\left(\sqrt{3}-1\right)\)
https://hoc24.vn/hoi-dap/question/405366.html
\(\sqrt{4-\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right)\left(4+\sqrt{15}\right)\)
\(=\sqrt{\left(4+\sqrt{15}\right)^2\left(4-\sqrt{15}\right)}\times\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{\left(4+\sqrt{15}\right)\left(16-15\right)}\times\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{8+2\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\left(\sqrt{5}-\sqrt{3}\right)\)
= 5 - 3
= 2
Bài 1:
Xét tử số:
\(\sqrt{14+6\sqrt{5}}-\sqrt{14-6\sqrt{5}}=\sqrt{3^2+5+2.3\sqrt{5}}-\sqrt{3^2+5-2.3\sqrt{5}}\)
\(=\sqrt{(3+\sqrt{5})^2}-\sqrt{(3-\sqrt{5})^2}=3+\sqrt{5}-(3-\sqrt{5})=2\sqrt{5}\)
Xét mẫu số:
\(\sqrt{(\sqrt{5}+1)\sqrt{6-2\sqrt{5}}}=\sqrt{(\sqrt{5}+1)\sqrt{5+1-2\sqrt{5}}}=\sqrt{(\sqrt{5}+1)\sqrt{(\sqrt{5}-1)^2}}\)
\(=\sqrt{(\sqrt{5}+1)(\sqrt{5}-1)}=\sqrt{4}=2\)
Do đó: $A=\frac{2\sqrt{5}}{2}=\sqrt{5}$
a: \(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
b: \(=\dfrac{\left(3-\sqrt{5}\right)\left(\sqrt{5}+1\right)+\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)}{\sqrt{2}}\)
\(=\dfrac{3\sqrt{5}+3-5-\sqrt{5}+3\sqrt{5}-3+5-\sqrt{5}}{\sqrt{2}}\)
\(=\dfrac{4\sqrt{5}}{\sqrt{2}}=2\sqrt{10}\)
\(A=\sqrt{3+\sqrt{5+2\sqrt{3}}.\sqrt{3-\sqrt{5+2\sqrt{3}}}}=\sqrt{\left(3^2\right)-\left(\sqrt{5+2\sqrt{3}}\right)^2}\)
\(=\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
\(B=\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\)
\(=\sqrt{4+2\sqrt{2}}.\sqrt{2^2-2-\sqrt{2}}=\sqrt{2}.\sqrt{2+\sqrt{2}}.\sqrt{2-\sqrt{2}}\)
\(=\sqrt{2}.\sqrt{4-2}=\sqrt{2}.\sqrt{2}=2\)
\(C=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)
\(=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2^2-\left(2+\sqrt{2+\sqrt{3}}\right)}\)
\(=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2-\sqrt{2+\sqrt{3}}}=\sqrt{2+\sqrt{3}}.\sqrt{2^2-\left(2+\sqrt{3}\right)}\)
\(=\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}=\sqrt{4-3}=1\)
\(D=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\sqrt{4+\sqrt{15}}.\sqrt{2}.\left(\sqrt{5}-\sqrt{3}\right).\sqrt{4+\sqrt{15}}.\sqrt{4-\sqrt{15}}\)
\(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}.\left(\sqrt{5}-\sqrt{3}\right).\sqrt{4^2-15}\)
\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=5-3=2\)
\(E=\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right).\sqrt{3-\sqrt{5}}\)
\(=\sqrt{3+\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{3+\sqrt{5}}\)
\(=2\sqrt{3-\sqrt{5}}+2\sqrt{3+\sqrt{5}}=\sqrt{2}\left(\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}\right)\)
\(=\sqrt{2}.\left(\sqrt{5}-1+\sqrt{5}+1\right)=2\sqrt{10}\)
B1:
1. \(\sqrt{12.5}\cdot\sqrt{0.2}\cdot\sqrt{0.1}\) \(=\sqrt{12.5\cdot0.2\cdot0.1}\) \(=\sqrt{0.25}=0.5\)
2.\(\sqrt{48.4}\cdot\sqrt{5}\cdot\sqrt{0.5}\) = \(\sqrt{48.4\cdot5\cdot0.5}\) =\(\sqrt{121}=11\)
B2:
a, \(\left(\sqrt{7}+\sqrt{3}\right)^2=7+2\cdot\sqrt{7}\cdot\sqrt{3}+3=7+2\cdot\sqrt{21}+3\)\(=10+2\sqrt{21}\)
b,\(\left(\sqrt{11}-\sqrt{5}\right)^2=11-2\sqrt{55}+5=16-2\sqrt{55}\)
c,\(\left(\sqrt{x}+\sqrt{y}\right) ^2=x+2\sqrt{xy}+y\)
d.\(\left(\sqrt{13}+\sqrt{7}\right)^2=13+2\sqrt{7}+7=20+2\sqrt{7}\)
e,\(\left(\sqrt{a}-\sqrt{b}\right)^2=a-2\sqrt{ab}+b\)
f,\(\left(\sqrt{3}-1\right)^2=3-2\sqrt{3}+1=4-2\sqrt{3}\)
\(=\sqrt{9\cdot\left(-2\right)^2\cdot\left(3+4+8\sqrt{3}\right)}\)
\(=3\cdot2\cdot\sqrt{7+8\sqrt{3}}\)
\(=6\sqrt{8\sqrt{3}+7}\)