Tính bằng cách thuận tiện\(\left(1-\frac{1}{2}\right)\div\)(\(1-\frac{1}{3}\))\(\div\)\(\left(1-\frac{1}{4}\right)\)
K
Khách
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30 tháng 4 2017
Ta có S = ( 1/2 - 1) : ( 1/3 - 1) : (1/4 - 1) :... : ( 1/50 - 1)
S = -1/2 : ( -2/3) : ( -3/4) : ... : ( -49/ 50)
S= -1/2 x (-3/2) x ( -4/3) x ... x (-50/49)
S= -1/2 x 1/3 x 50
S= -25/3
KN
25 tháng 8 2020
\(4.\left(\frac{1}{4}\right)^2+25\left[\left(\frac{3}{4}\right)^3:\left(\frac{5}{4}\right)^3\right]:\left(\frac{3}{2}\right)^3=4.\frac{1}{16}+25\left(\frac{27}{64}.\frac{64}{125}\right).\frac{8}{27}\)
\(=\frac{1}{4}+25.\frac{27}{125}.\frac{8}{27}=\frac{1}{4}+\frac{8}{5}=\frac{37}{20}\)
\(2^3+3\left(\frac{1}{2}\right)^0-1+\left[\left(-2\right)^2:\frac{1}{2}\right]-8=8+3-1+4.2-8=10\)
22 tháng 6 2016
=\(2\left(\frac{1}{2}-\frac{1}{2.3}\right).2\left(\frac{1}{2}-\frac{1}{3.4}\right)....2\left(\frac{1}{2}-\frac{1}{99.100}\right)\)
=\(2^{89}\left(\frac{1}{2}.98-\frac{1}{2}+\frac{1}{100}\right)\)
\(=2^{98}.\left(49-\frac{49}{100}\right)=\frac{2^{98}.4851}{100}\)
BH
19 tháng 8 2019
\(a,\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=\frac{1}{2}-\frac{1}{6}=\frac{1}{3}\)
\(b,\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\)
\(=\frac{1\times2\times3}{2\times3\times4}=\frac{1}{4}\)
PA
8 tháng 9 2016
\(\left[\left(1+\frac{1}{x^2}\right)\div\left(1+2x+x^2\right)+\frac{2}{\left(x+1\right)^3}\times\left(1+\frac{1}{x}\right)\right]\div\frac{x-1}{x^3}\)
\(=\left[\frac{x^2+1}{x^2}\times\frac{1}{\left(x+1\right)^2}+\frac{2}{\left(x+1\right)^3}\times\frac{x+1}{x}\right]\div\frac{x-1}{x^3}\)
\(=\left(\frac{x^2+1}{x^2}\times\frac{1}{\left(x+1\right)^2}+\frac{1}{\left(x+1\right)^2}\times\frac{2}{x}\right)\div\frac{x-1}{x^3}\)
\(=\left(\frac{1}{\left(x+1\right)^2}\times\left(\frac{x^2+1}{x^2}+\frac{2}{x}\right)\right)\div\frac{x-1}{x^3}\)
\(=\left(\frac{1}{\left(x+1\right)^2}\times\frac{x^3+2x^2+x}{x^3}\right)\div\frac{x-1}{x^3}\)
\(=\left(\frac{1}{\left(x+1\right)^2}\times\frac{x\left(x^2+2x+1\right)}{x^3}\right)\div\frac{x-1}{x^3}\)
\(=\left(\frac{1}{\left(x+1\right)^2}\times\frac{x\left(x+1\right)^2}{x^3}\right)\div\frac{x-1}{x^3}\)
\(=\frac{1}{x^2}\times\frac{x^3}{x-1}\)
\(=\frac{x}{x-1}\)
14 tháng 10 2016
Bấm máy tính:
E = \(\frac{4}{3}+\frac{1}{4}+\frac{3}{5}:\frac{4}{5}\)
E = \(\frac{4}{3}+\frac{1}{4}+\frac{3}{4}\)
E = \(\frac{7}{3}\)
Vậy E = \(\frac{7}{3}\)
15 tháng 10 2016
\(E=\frac{0,8:\left(\frac{4}{5}.1,25\right)}{0,64-\frac{1}{25}}+\frac{\left(1,08-\frac{2}{25}\right):\frac{4}{7}}{\left(6\frac{5}{9}-3\frac{1}{4}\right).2\frac{2}{17}}+\left(1,2.0,5\right):\frac{4}{5}\)
\(E=\frac{\frac{4}{5}:\frac{4}{5}:1,25}{\frac{16}{25}-\frac{1}{25}}+\frac{\left(\frac{27}{25}-\frac{2}{25}\right).\frac{7}{4}}{\left(\frac{59}{9}-\frac{13}{4}\right).\frac{36}{17}}+\frac{6}{5}.\frac{1}{2}.\frac{5}{4}\)
\(E=\frac{1:\frac{5}{4}}{\frac{3}{5}}+\frac{1.\frac{7}{4}}{\frac{119}{36}.\frac{36}{17}}+\frac{3}{4}\)
\(E=\frac{4}{5}.\frac{5}{3}+\frac{\frac{7}{4}}{7}+\frac{3}{4}\)
\(E=\frac{4}{3}+\frac{7}{4}.\frac{1}{7}+\frac{3}{4}\)
\(E=\frac{4}{3}+\frac{1}{4}+\frac{3}{4}\)
\(E=\frac{4}{3}+1=\frac{7}{3}\)
(1-1/2):(1-1/3):(1-1/4)
=1/2:2/3:3/4
=1/2 x 3/2 x 4/3
=1 x 3 x 4/ 2 x 2 x 3
=3 x 4/4 x 3
=12/12
=1