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AL
3 tháng 8 2016
\(\frac{1,4.2012.6+4,2.4024+8,4.976}{1+2+3+4+...+999}\)
\(=\frac{8,4.2012+4,2.2.2012+8,4.976}{1+2+3+4+...+999}\)
\(=\frac{8,4\left(2012+2012+976\right)}{1+2+3+4+...+999}\)
\(=\frac{8,4.5000}{500.999}\)
\(=\frac{84.500}{999.500}\)
\(=\frac{84}{999}\)
\(=\frac{28}{333}\)
TT
7
20 tháng 12 2016
\(a,27.75+27.25\)
\(=27.\left(75+25\right)\)
\(=27.100\)
\(=2700\)
\(b,\left(-6\right)+20+\left(-4\right)\)
\(=\left(-6-4\right)+20\)
\(=-10+20\)
\(=10\)
\(c,2^2.3^1-\left(1^{2012}+2012^0\right):2\)
\(=4.3-\left(1+1\right):2\)
\(=12-2:2\)
\(=12-1\)
\(=11\)
20 tháng 12 2016
a)27.75+27.25
=27.(75+25)
=27.100
=2700
b)(-6)+20+(-4)
=[(-6)+(-4)]+20
=(-10)+20
=10
c)22.3-(12012+20120):2
=12-2:2
=12-1
=11
11 tháng 7 2016
\(\frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\frac{2013}{1}+\frac{2014}{2}+\frac{2015}{3}+...+\frac{4024}{2012}-2012}\)
\(=\frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\left(\frac{2013}{1}-1\right)+\left(\frac{2014}{2}-1\right)+\left(\frac{2015}{3}-1\right)+...+\left(\frac{4024}{2012}-1\right)}\)
\(=\frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\frac{2012}{1}+\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2012}}\)
\(=\frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{2012.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}\right)}\)
\(=\frac{1}{2012}\)
Ủng hộ mk nha ^_-
DT
0
12 tháng 8 2020
\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)
=> \(2A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\right)\)
=> \(A=2-\frac{1}{2^{2012}}=\frac{2^{2013}-1}{2^{2012}}\)
12 tháng 8 2020
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
\(2A=2\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)
\(2A=3+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2011}}\)
\(2A-A=A\)
\(=\left(3+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)
\(=3+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2011}}-1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{2012}}\)
\(=2-\frac{1}{2012^2}\)
\(B=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{9999}\right)\cdot\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
\(B=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{9999}\right)\cdot\left(\frac{6}{12}-\frac{4}{12}-\frac{2}{12}\right)\)
\(B=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{9999}\right)\cdot0=0\)
