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24 tháng 8 2017
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) \(\left(\sqrt{3}-\sqrt{2}\right)^2=3-2.\sqrt{3}.\sqrt{2}+2=5-2\sqrt{6}\)
\(\left(3\sqrt{2}-1\right)\left(3\sqrt{2}+1\right)=3\sqrt{2}.3\sqrt{2}+3\sqrt{2}-3\sqrt{2}-1=18-1\)
\(\left(\sqrt{6}+2\right)\left(\sqrt{3}-2\right)=\sqrt{6}.\sqrt{3}-2\sqrt{6}+2\sqrt{3}-4=\sqrt{18}-2\sqrt{6}+2\sqrt{3}-4\)\(=3\sqrt{2}-2\sqrt{6}+2\sqrt{3}-4\)
\(c,\left(\sqrt{\dfrac{3}{2}}-\sqrt{\dfrac{2}{3}}\right)=\dfrac{\sqrt{3}}{\sqrt{2}}-\dfrac{\sqrt{2}}{\sqrt{3}}=\dfrac{3-2}{\sqrt{2}\sqrt{3}}\) = \(\dfrac{1}{\sqrt{6}}\)
\(\left(\sqrt{\dfrac{8}{3}}-\sqrt{24}+\sqrt{\dfrac{50}{3}}\right).\sqrt{6}=\sqrt{\dfrac{8}{3}}.\sqrt{6}-\sqrt{24}.\sqrt{6}+\sqrt{\dfrac{50}{3}}.\sqrt{6}\) = \(\dfrac{\sqrt{8}.\sqrt{6}}{\sqrt{3}}-\sqrt{144}+\dfrac{\sqrt{50}.\sqrt{6}}{\sqrt{3}}=\dfrac{\sqrt{48}}{\sqrt{3}}-12+\dfrac{\sqrt{300}}{\sqrt{3}}=\sqrt{\dfrac{48}{3}}-12+\sqrt{\dfrac{300}{3}}=4-12+10=2\)
24 tháng 8 2017
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) \(\sqrt{\left(\sqrt{2}-3\right)^2}.\sqrt{11+6\sqrt{2}}=\left|\sqrt{2}-3\right|.\sqrt{2+6\sqrt{2}+9}=\left(\sqrt{2}-3\right).\sqrt{\left(\sqrt{2}+3\right)^2}=\left(\sqrt{2}-3\right)\)\(\left(\sqrt{2}+3\right)=2+3\sqrt{2}-3\sqrt{2}-9=-7\)
\(\sqrt{\left(\sqrt{3}-3\right)^2}.\sqrt{\dfrac{1}{3-\sqrt{3}}}=\left|\sqrt{3}-3\right|.\dfrac{1}{3-\sqrt{3}}=-\left(3-\sqrt{3}\right).\left(\dfrac{1}{3-\sqrt{3}}\right)=-1\)
12 tháng 10 2020
a) \(\left(\frac{\sqrt{9}}{2}+\frac{\sqrt{1}}{2}-\sqrt{2}\right)\sqrt{2}\)
\(=\frac{3\sqrt{2}}{2}+\frac{\sqrt{2}}{2}-2\)
\(=\frac{4\sqrt{2}}{2}-2=2\sqrt{2}-2\)
b) \(\left(\frac{\sqrt{8}}{3}-\sqrt{24}+\frac{\sqrt{50}}{3}\right)\sqrt{6}\)
\(=\frac{4\sqrt{3}}{3}-12+\frac{10\sqrt{3}}{3}\)
\(=\frac{14\sqrt{3}}{3}-12\)
c) \(\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{1}\right)\) (đã sửa đề)
\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\sqrt{2}\)
\(=\left(3-1\right)\sqrt{2}\)
\(=2\sqrt{2}\)
d) \(\left(3\sqrt{2}+1\right)\left(\sqrt{3\sqrt{2}-1}\right)\)
\(=\sqrt{3\sqrt{2}+1}\cdot\left(\sqrt{3\sqrt{2}+1}\cdot\sqrt{3\sqrt{2}-1}\right)\)
\(=\sqrt{3\sqrt{2}+1}\cdot\sqrt{18-1}\)
\(=\sqrt{3\sqrt{2}+1}\cdot\sqrt{17}\)
...
HP
10 tháng 10 2020
a, \(=2\sqrt{7}-8+15\sqrt{7}-12=17\sqrt{7}-20\)
b, \(=2\sqrt{2}-10\sqrt{2}+4\sqrt{2}=-4\sqrt{2}\)
c, \(=\frac{3}{8}.\frac{4}{3}-2.\frac{2}{5}=\frac{1}{2}-\frac{4}{5}=-\frac{3}{10}\)
d, \(\sqrt{\left(\sqrt{3-1}\right)^2}-\sqrt{\left(\sqrt{3-2}\right)^2}=\sqrt{3-1}-\sqrt{3-2}=\sqrt{2}-\sqrt{1}=\sqrt{2}-1\)
e, \(\sqrt{2-3}\) không tồn tại
4 tháng 7 2018
\(1.A=\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)=5-4=1\)
\(2.B=\left(\sqrt{45}+\sqrt{63}\right)\left(\sqrt{7}-\sqrt{5}\right)=\left(3\sqrt{5}+3\sqrt{7}\right)\left(\sqrt{7}-\sqrt{5}\right)=2\left(7-5\right)=4\) \(3.C=\left(\sqrt{5}+\sqrt{3}\right)\left(5-\sqrt{15}\right)=\sqrt{5}\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=\sqrt{5}\left(5-3\right)=2\sqrt{5}\) \(4.\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)=\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)=\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)=27-2=25\) \(5.E=\left(\sqrt{3}+1\right)^2-2\sqrt{3}+4=4+2\sqrt{3}-2\sqrt{3}+4=8\)
\(6.F=\left(\sqrt{15}-2\sqrt{3}\right)^2+12\sqrt{5}=27-12\sqrt{5}+12\sqrt{5}=27\)
29 tháng 9 2017
a) \(\sqrt{18}\)-2\(\sqrt{50}\)+\(\sqrt{\left(2-\sqrt{2}\right)^2}\)
=3\(\sqrt{2}\)-10\(\sqrt{2}\)+(2-\(\sqrt{2}\))2
= 3\(\sqrt{2}\)-10\(\sqrt{2}\)+4-2
= -7\(\sqrt{2}\)+2
LD
29 tháng 9 2017
a) \(\sqrt{18}-2\sqrt{50}+\sqrt{\left(2-\sqrt{2}\right)^2}\)
=\(3\sqrt{2}-10\sqrt{2}+2-\sqrt{2}=2-8\sqrt{2}\)
b)\(\sqrt{\dfrac{1}{3}}+\dfrac{3}{\sqrt{3}}+\dfrac{1}{2-\sqrt{3}}\)
=\(\dfrac{1}{3}\sqrt{3}+\sqrt{3}+\dfrac{1}{2-\sqrt{3}}=\dfrac{4}{3}\sqrt{3}+\dfrac{1}{2-\sqrt{3}}\)
=\(\dfrac{4\sqrt{3}+2+\sqrt{3}}{3}=\dfrac{5\sqrt{3}+2}{3}\)
c)\(\left(1+\sqrt{2}+\sqrt{3}\right)\left(1+\sqrt{2}-\sqrt{3}\right)\)
=\(\left(1+\sqrt{2}\right)^2-3=1+2\sqrt{2}+2-3=2\sqrt{2}\)
d)\(3\sqrt{200}-2\sqrt{0,08}-4\sqrt{\dfrac{9}{8}}\)
=\(30\sqrt{2}-0,4\sqrt{2}-3\sqrt{2}=26.6\sqrt{2}\)
15 tháng 7 2019
\(\frac{3\sqrt{128}}{\sqrt{2}}=\frac{\sqrt{9.128}}{\sqrt{2}}=\sqrt{\frac{1152}{2}}=\sqrt{576}=24\)
\(A=3\sqrt{8}-\sqrt{50}-\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(=3\sqrt{2.2^2}-\sqrt{2.5^2}-\left(\sqrt{2}-1\right)\)
\(=6\sqrt{2}-5\sqrt{2}-\sqrt{2}+1\)
\(=\sqrt{2}\left(6-5-1\right)+1\)
\(=\sqrt{2}.0+1\)
\(=1\)