tính nhanh: 2011.2013-20122
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\(A=\left(2012-1\right)\left(2012+1\right)-2012^2=2012^2-1-2012^2=-1\)
\(A=\left(2012-1\right)\left(2012+1\right)-2012^2=2012^2-1-2012^2=-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}\left(3^{32}-1\right)< 3^{32}-1=B\)
b) \(A=2011.2013=\left(2012-1\right)\left(2012+1\right)=2012^2-1< 2012^2=B\)
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[(23 - 5) . (-3)+9] . (22012 . 2011 - 20122 . 2011+1)
= [ 3 . ( -3 ) + 9] . (22012 . 2011 - 20122 . 2011+1)
= [ (-9) + 9 ] . (22012 . 2011 - 20122 . 2011+1)
= 0 . (22012 . 2011 - 20122 . 2011+1)
= 0
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\(A=2011.2013-2012^2\)
Gọi 2012 là a ta có:
\(2011=a-1;2013=a+1\)
\(\Rightarrow A=\left(a+1\right).\left(a-1\right)-a^2\)
\(\Rightarrow A=a^2-a+a-1-a^2\)
\(\Rightarrow A=a^2-1-a^2\)
\(\Rightarrow A=-1\)
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\(M=\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{2011.2013}\)
\(M=2.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2011.2013}\right)\)
\(M=2.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2011}-\frac{1}{2013}\right)\)
\(M=2.\left(1-\frac{1}{2013}\right)\)
\(M=2.\frac{2012}{2013}\)
\(M=\frac{4024}{2013}\)
~Học tốt~
M=2.(2/1.3 + 2/3.5 +2/5.7+...+2/2011.2013)
M=2.(1-1/3 +1/3-1/5 +1/5-1/7+... +1/2011-1/2013)
M=2.(1-1/2013)
M=2.2012/2013
M=4024/2013
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(\frac{4}{1.3}\)+\(\frac{4}{3.5}\)+\(\frac{4}{5.7}\)+\(\frac{4}{7.9}\)+...+\(\frac{4}{2011.2013}\)
= 1+\(\frac{1}{3}\)-\(\frac{1}{3}\)+\(\frac{1}{5}\)-\(\frac{1}{5}\)+\(\frac{1}{7}\)-\(\frac{1}{7}\)+\(\frac{1}{9}\)+...+\(\frac{1}{2011}\)+\(\frac{1}{2013}\)
=1+ 0 + 0 + 0 +...+ 0 + \(\frac{1}{2013}\)
=1+\(\frac{1}{2013}\)
=\(\frac{2014}{2013}\)
k dùm nha
\(\frac{4}{1\cdot3}+\frac{4}{3\cdot5}+\frac{4}{5\cdot7}+...+\frac{4}{2011\cdot2013}\)
\(=2\cdot\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{2011\cdot2013}\right)\)
\(=2\cdot\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2011}-\frac{1}{2013}\right)\)
\(=2\cdot\left(1-\frac{1}{2013}\right)\)
\(=2\cdot\frac{2012}{2013}\)
\(=\frac{4024}{2013}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a = 2011.2013
a = 2011.(2012+1)
a = 2011.2012 + 2011
b = 2012.2012
b = (2011+1).2012
b = 2011.2012 + 2012
Vì 2011 < 2012
=> 2011.2012 + 2011 < 2011.2012 + 2012
=> a < b
2011.2013-2012^2=(2012-1).(2012+1)-2012^2
=2012^2-1^2-2012^2
=2012^2-2012^2-1^2
=-1