B= 4/1\(\times\)3+4/3\(\times\)5+4/5\(\times\)7+...|+4/47\(\times\)49+4/49\(\times\)51
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MV
23 tháng 5 2017
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\\ 2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\\ 2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right)\\ A=1-\dfrac{1}{2^{100}}\)
MV
23 tháng 5 2017
\(E=\dfrac{3^2}{2\cdot4}+\dfrac{3^2}{4\cdot6}+...+\dfrac{3^2}{198\cdot200}\\ =3^2\cdot\left(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{198\cdot200}\right)\\ =9\cdot\dfrac{1}{2}\cdot\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{198\cdot200}\right)\\ =\dfrac{9}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{198}-\dfrac{1}{200}\right)\\ =\dfrac{9}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{200}\right)\\ =\dfrac{9}{2}\cdot\dfrac{99}{200}\\ =\dfrac{891}{400}\)
6 tháng 8 2018
\(\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}\right).\dfrac{1-3-5-7-...-49}{89}\)
\(=\dfrac{1}{5}.\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{19}+...+\dfrac{1}{44}-\dfrac{1}{99}\right).\dfrac{1-\left(3+5+...+49\right)}{89}\)
\(=\dfrac{1}{5}.\left(\dfrac{1}{4}-\dfrac{1}{49}\right).\dfrac{1-\dfrac{\left(3+49\right).\left[\left(49-3:2+1\right)\right]}{2}}{89}\)
\(=\dfrac{1}{5}.\dfrac{45}{196}.\dfrac{1-\dfrac{52.24}{2}}{89}\)
\(=\dfrac{9}{196}.\dfrac{1-\dfrac{1248}{2}}{89}\)
\(=\dfrac{9}{196}.\dfrac{1-624}{89}\)
\(=\dfrac{9}{196}.\left(-\dfrac{623}{89}\right)\)
\(=-\dfrac{9}{28}\)
20 tháng 2 2020
a) Ta có: \(14-\left(5-8\right)^3+\left(-3\right)\cdot5\)
\(=14+27-15=26\)
b) Ta có: \(-7\cdot15+7\cdot\left(-35\right)+\left(-1\right)^{2019}\)
\(=-7\left(15+35\right)-1=-350-1=-351\)
c) Ta có: \(-18\cdot32+\left(-18\right)\cdot45+77\cdot\left(-32\right)-77\cdot50\)
\(=-18\left(32+45\right)+77\left(-32-50\right)\)
\(=-18\cdot77+77\cdot\left(-82\right)=77\left(-18-82\right)=77\cdot\left(-100\right)=-7700\)
d) Ta có: \(104-4\cdot\left[-5\cdot8+\left(7-10\right)^3\right]\)
\(=104-4\left[-40-27\right]=104-4\cdot\left(-67\right)=104+268=372\)
VM
13 tháng 8 2019
3.
a) \(\left(x-1\right)^3=125\)
=> \(\left(x-1\right)^3=5^3\)
=> \(x-1=5\)
=> \(x=5+1\)
=> \(x=6\)
Vậy \(x=6.\)
b) \(2^{x+2}-2^x=96\)
=> \(2^x.\left(2^2-1\right)=96\)
=> \(2^x.3=96\)
=> \(2^x=96:3\)
=> \(2^x=32\)
=> \(2^x=2^5\)
=> \(x=5\)
Vậy \(x=5.\)
c) \(\left(2x+1\right)^3=343\)
=> \(\left(2x+1\right)^3=7^3\)
=> \(2x+1=7\)
=> \(2x=7-1\)
=> \(2x=6\)
=> \(x=6:2\)
=> \(x=3\)
Vậy \(x=3.\)
Chúc bạn học tốt!
\(B=\dfrac{4}{1\times3}+\dfrac{4}{3\times5}+\dfrac{4}{5\times7}+...+\dfrac{4}{47\times49}+\dfrac{4}{49\times51}\)
\(=2\times\left(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+...+\dfrac{2}{47\times49}+\dfrac{2}{49\times51}\right)\)
\(=2\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{47}-\dfrac{1}{49}+\dfrac{1}{49}-\dfrac{1}{51}\right)\)
\(=2\times\left(1-\dfrac{1}{51}\right)\)
\(=2\times\dfrac{50}{51}\)
\(=\dfrac{100}{51}\)