\(\sqrt{3-\sqrt{5}}\) * \(\sqrt{3+...">
K
Khách

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14 tháng 7 2016

\(A=\sqrt{3+\sqrt{5}}.\sqrt{3-\sqrt{5}}=\sqrt{3^2-\left(\sqrt{5}\right)^2}=\sqrt{4}=2\)

\(B=\sqrt{150.27.96}=\sqrt{150}.\sqrt{27}.\sqrt{96}=5\sqrt{6}.3\sqrt{3}.4\sqrt{6}=360\sqrt{3}\)

\(C=\left(\sqrt{27}+\sqrt{48}\right)^2-\left(\sqrt{27}-\sqrt{48}\right)^2\)\(=\left[\left(\sqrt{27}+\sqrt{48}-\sqrt{27}+\sqrt{48}\right)\left(\sqrt{27}+\sqrt{48}+\sqrt{27}-\sqrt{48}\right)\right]\)

\(=2\sqrt{27}.2\sqrt{48}=2.3\sqrt{3}.2.4\sqrt{3}=144\)

\(D=\sqrt{137^2-88^2}-\sqrt{192^2-111^2}=\sqrt{\left(137+88\right)\left(137-88\right)}-\sqrt{\left(192+111\right)\left(192-111\right)}\)

\(=\sqrt{225.49}-\sqrt{303.81}=15.7-9.\sqrt{303}=9\left(\frac{35}{3}-\sqrt{303}\right)\)

14 tháng 7 2016

\(E=\sqrt{\frac{225}{4}.\frac{81}{25}.\frac{49}{64}}=\frac{15}{2}.\frac{9}{5}.\frac{7}{8}=\frac{189}{16}\)

\(F=\sqrt{\frac{27}{25}}.\sqrt{\frac{49}{189}}.\sqrt{\frac{700}{99}}=\frac{3\sqrt{3}}{5}.\frac{7}{3\sqrt{21`}}.\frac{10\sqrt{7}}{3\sqrt{11}}=\frac{14}{3\sqrt{11}}\)

\(H=\sqrt{105}.\left[\sqrt{\frac{15}{7}}-\sqrt{\frac{35}{5}}+\sqrt{\frac{21}{5}}\right]=\sqrt{105}.\left[\sqrt{\frac{15}{7}}-\sqrt{7}+\sqrt{\frac{21}{5}}\right]\)

\(=\sqrt{105}.\left[\frac{\sqrt{75}-\sqrt{49}+\sqrt{147}}{\sqrt{35}}\right]=\sqrt{3}\left(12\sqrt{3}-7\right)=36-7\sqrt{3}\)

\(K=\sqrt{64.14.21.54}-\sqrt{35.45.12}=8.\sqrt{14}.\sqrt{21}.3\sqrt{6}-\sqrt{35}.3\sqrt{5}.2\sqrt{3}\)

\(=24.\sqrt{14.21.6}-6\sqrt{35.5.3}=24.42-30\sqrt{21}=30\left(\frac{168}{5}-\sqrt{21}\right)\)

AH
Akai Haruma
Giáo viên
30 tháng 6 2019

a)

\((\sqrt{3}-2\sqrt{12}+2\sqrt{4})(\sqrt{27}+\sqrt{144}-2\sqrt{16})\)

\(=(\sqrt{3}-4\sqrt{3}+4)(3\sqrt{3}+12-8)\)

\(=(-3\sqrt{3}+4)(3\sqrt{3}+4)=4^2-(3\sqrt{3})^2=16-27=-11\)

b)

\((2\sqrt{5}+2\sqrt{3})^2-4\sqrt{60}\)

\(=(2\sqrt{5})^2+2.2\sqrt{5}.2\sqrt{3}+(2\sqrt{3})^2-8\sqrt{15}\)

\(=32+8\sqrt{15}-8\sqrt{15}=32\)

c)

\(\sqrt{6}(3\sqrt{12}-4\sqrt{3}+\sqrt{48}-5\sqrt{6})\)

\(=3\sqrt{72}-4\sqrt{18}+\sqrt{6.48}-5.\sqrt{36}\)

\(=18\sqrt{2}-12\sqrt{2}+12\sqrt{2}-30=18\sqrt{2}-30\)

AH
Akai Haruma
Giáo viên
30 tháng 6 2019

d)

\((\sqrt{2}-\sqrt{3})(\sqrt{6}+\sqrt{2})(\sqrt{2}+\sqrt{3})\)

\(=(\sqrt{2}-\sqrt{3})(\sqrt{2}+\sqrt{3})(\sqrt{6}+\sqrt{2})\)

\(=(2-3)(\sqrt{6}+\sqrt{2})=-(\sqrt{6}+\sqrt{2})\)

e) Biểu thức bên trong căn lớn âm nên biểu căn bậc 2 không có nghĩa

f)

\((\frac{2}{\sqrt{3}-1}+\frac{3}{\sqrt{3}-2}+\frac{15}{3-\sqrt{3}}).\frac{1}{\sqrt{3}+5}\)

\(=(\frac{2\sqrt{3}+15}{3-\sqrt{3}}+\frac{3}{\sqrt{3}-2}).\frac{1}{\sqrt{3}+5}\)

\(=\frac{2\sqrt{3}+15)(\sqrt{3}-2)+3(3-\sqrt{3})}{(3-\sqrt{3})(\sqrt{3}-2)}.\frac{1}{\sqrt{3}+5}\)

\(=\frac{-15+8\sqrt{3}}{(-9+5\sqrt{3})(\sqrt{3}+5)}=\frac{-15+8\sqrt{3}}{-30+16\sqrt{3}}=\frac{-15+8\sqrt{3}}{2(-15+8\sqrt{3})}=\frac{1}{2}\)

28 tháng 6 2019

a)(\(\sqrt{2006}-\sqrt{2005}\)).(\(\sqrt{2006}+\sqrt{2005}\))

=\(\sqrt{2006}^2-\sqrt{2005}^2\)

=2006-2005

=1

\(a,\sqrt{\frac{72}{9}}:\sqrt{8}=\frac{\sqrt{72}}{\sqrt{9}}.\frac{1}{\sqrt{8}}\)

\(=\frac{6\sqrt{2}}{3}.\frac{1}{2\sqrt{2}}\)

\(=1\)

\(b,\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right):\sqrt{3}=\left(28\sqrt{3}+9\sqrt{3}-4\sqrt{3}\right):\sqrt{3}\)

\(=33\sqrt{3}:\sqrt{3}\)

\(=33\)

\(c,\left(\sqrt{125}+\sqrt{245}-\sqrt{5}\right):\sqrt{5}=\left(5\sqrt{5}+7\sqrt{5}-\sqrt{5}\right):\sqrt{5}\)

\(=11\sqrt{5}:\sqrt{5}\)

\(=11\)

\(d,\left(\sqrt{\frac{1}{7}}-\sqrt{\frac{16}{7}}+\sqrt{7}\right):\sqrt{7}=\left(\frac{1}{\sqrt{7}}-\frac{4}{\sqrt{7}}+\frac{7}{\sqrt{7}}\right):\sqrt{7}\)

\(=\frac{4}{\sqrt{7}}.\frac{1}{\sqrt{7}}=\frac{4}{7}\)

1 tháng 4 2020

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 tháng 4 2020

ban ơi ccachs làm

22 tháng 7 2017

a, \(\frac{1}{\left(\sqrt{3}+\sqrt{2}\right)^2}\) +\(\frac{1}{\left(\sqrt{3}-\sqrt{2}\right)^2}\) =\(\frac{\left(\sqrt{3}+\sqrt{2}\right)^2+\left(\sqrt{3}-\sqrt{2}\right)^2}{\left(\sqrt{3}+\sqrt{2}\right)^2\left(\sqrt{3}-\sqrt{2}\right)^2}\) 

                                                                         \(=\frac{10}{1}=10\)

mấy câu còn lại bạn tự làm nốt nhé mk ban rồi 

22 tháng 7 2017

Câu bạn trả lời ở đâu v 

21 tháng 7 2017

1,\(4\sqrt{5}+2\sqrt{5}-\sqrt{5}-15\sqrt{5}=-10\sqrt{5}\)

2,\(8\sqrt{5}-15\sqrt{5}+15\sqrt{5}-3\sqrt{5}=5\sqrt{5}\)

3,\(\left(28\sqrt{3}+9\sqrt{3}-4\sqrt{3}\right):\sqrt{3}=33\) 

4,\(7\sqrt{7a}+3\sqrt{7a}-2\sqrt{7a}=8\sqrt{7a}\)

5,\(-6\sqrt{a}-\sqrt{6a}+\sqrt{6a}=-6\sqrt{a}\)

6,\(8\sqrt{3}-12\sqrt{3}+5\sqrt{3}+2\sqrt{3}=3\sqrt{3}\)