\(\left(3\sqrt{8}-6\sqrt{\frac{1}{2}}-2\sqrt{18}+3\sqrt{50}\right)\div\left...">
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28 tháng 7 2016

Hỏi đáp Toán

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) \(\left(3\sqrt{2}+1\right)\left(\sqrt{3\sqrt{2}-1}\right)\)

\(=\sqrt{3\sqrt{2}+1}\cdot\left(\sqrt{3\sqrt{2}+1}\cdot\sqrt{3\sqrt{2}-1}\right)\)

\(=\sqrt{3\sqrt{2}+1}\cdot\sqrt{18-1}\)

\(=\sqrt{3\sqrt{2}+1}\cdot\sqrt{17}\)

...

\(a,\sqrt{\left(\sqrt{2}-3\right)^2}.\sqrt{11+6\sqrt{2}}\)

\(=|\sqrt{2}-3|.\sqrt{9+6\sqrt{2}+2}\)

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

\(=\left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right)\)

\(=9-2=7\)

\(b,\sqrt{\left(\sqrt{3}-3\right)^2}.\sqrt{\frac{1}{3-\sqrt{3}}}\)

\(=\left(3-\sqrt{3}\right).\frac{\sqrt{1}}{\sqrt{3-\sqrt{3}}}\)

\(=\frac{3-\sqrt{3}}{\sqrt{3-\sqrt{3}}}\)

\(=\sqrt{3-\sqrt{3}}\)

\(c,-\frac{2}{3}\sqrt{\frac{\left(a-b\right)^3.b^5}{c}}.\frac{9}{4}\sqrt{\frac{c^3}{2\left(a-b\right)}}.\sqrt{98b}\)

\(=-\frac{2}{3}.\frac{\sqrt{\left(a-b\right)^3.b^5}}{\sqrt{c}}.\frac{9}{4}.\frac{\sqrt{c^3}}{\sqrt{2\left(a-b\right)}}.7\sqrt{2b}\)

\(=-\frac{2}{3}.\frac{\left(a-b\right)b^2\sqrt{\left(a-b\right)b}}{\sqrt{c}}.\frac{9}{4}.\frac{c\sqrt{c}}{\sqrt{2\left(a-b\right)}}.7\sqrt{2b}\)

\(=-\frac{2}{3}.\frac{9}{4}.7.\frac{\left(a-b\right).b^2\sqrt{\left(a-b\right)b}}{\sqrt{c}}.\frac{c\sqrt{c}}{\sqrt{2\left(a-b\right)}}.\sqrt{2b}\)

\(=-\frac{21}{2}.\left(a-b\right).b^2\sqrt{b}.c.\sqrt{b}\)

\(=\frac{-21}{2}.\left(a-b\right).b^3.c\)

\(d,\left(\sqrt{6}-3\sqrt{3}+5\sqrt{2}-\frac{1}{2}\sqrt{8}\right).2\sqrt{6}\)

\(=\left(\sqrt{6}-3\sqrt{3}+5\sqrt{2}-\frac{1}{2}.2\sqrt{2}\right).2\sqrt{6}\)

\(=\left(\sqrt{6}-3\sqrt{3}+5\sqrt{2}-\sqrt{2}\right).2\sqrt{6}\)

\(=\left(\sqrt{6}-3\sqrt{3}+4\sqrt{2}\right).2\sqrt{6}\)

\(=2.6-18\sqrt{2}+16\sqrt{3}\)

\(=12-18\sqrt{2}+16\sqrt{3}\)

28 tháng 7 2016

\(\left(\sqrt{4,5}-\frac{1}{2}.\sqrt{72}+5\sqrt{\frac{1}{2}}\right).\left(42\sqrt{\frac{25}{6}}-10\sqrt{\frac{3}{2}}-12\sqrt{\frac{98}{3}}\right)\)

=\(\left(\frac{3\sqrt{2}}{2}-3\sqrt{2}+\frac{5\sqrt{2}}{2}\right).\left(35\sqrt{6}-5\sqrt{6}-28\sqrt{6}\right)\)

=\(\left(\frac{3\sqrt{2}-6\sqrt{2}+5\sqrt{2}}{2}\right).2\sqrt{6}\)

=\(2\sqrt{2}.\sqrt{6}=4\sqrt{3}\)

7 tháng 9 2019

22) \(\frac{1}{\sqrt{5}+\sqrt{2}}+\frac{1}{\sqrt{5}-\sqrt{2}}\)

\(=\frac{\left(\sqrt{5}-\sqrt{2}\right)+\left(\sqrt{5}+\sqrt{2}\right)}{\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)}\)

\(=\frac{2\sqrt{5}}{\sqrt{5^2}-\sqrt{2^2}}\)

\(=\frac{2\sqrt{5}}{5-2}=\frac{2\sqrt{5}}{3}\)

11 tháng 10 2020

a) 2√2 -2

b) -12-√12

c) 3√6 -6

d) 17

e) -7

g) (29-6√12)/18

h) 2/3

l) (-2+5√5)/5

i) 1/6

k) 4

m) (√6+3)/3

n) 1

12 tháng 10 2020

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