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\(A=\left(2012-1\right)\left(2012+1\right)-2012^2=2012^2-1-2012^2=-1\)
\(A=\left(2012-1\right)\left(2012+1\right)-2012^2=2012^2-1-2012^2=-1\)
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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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\(A=2011.2013-2012^2\)
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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2011.2013+2012.2014
=(2013-2).2013+2012.(2012+2)
=20132-4026+20122+4024
=20132+20122+(-4026+4024)
=20132+20122-2
Ta có:\(2011.2013+2012.2014\)
\(=\left(2013-2\right).2013+\left(2012+2\right).2012\)
\(=2013^2-4026+2012^2+4024\)
\(=2012^2+2013^2-2\)
nên hai phép tính trên bằng nhau.
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=\dfrac{1^2}{1.3}+\dfrac{2^2}{3.5}+\dfrac{3^3}{5.7}+...+\dfrac{1006^2}{2011.2013}\)
\(\Rightarrow4A=\dfrac{4.1^2}{1.3}+\dfrac{4.2^2}{3.5}+\dfrac{4.3^3}{5.7}+...+\dfrac{4.1006^2}{2011.2013}\)
\(\Rightarrow4A=1006+\dfrac{1}{2}.\left[1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-...+\dfrac{1}{2011}-\dfrac{1}{2013}\right]\)
\(\Rightarrow A=\dfrac{1006+\dfrac{1}{2}.\left(1-\dfrac{1}{2013}\right)}{4}\)
\(\Rightarrow A=251,6249\)
Cho mik hỏi tại sao lại tính được đến 1006+1/2(1-1/3....) vấn đề của mik là số 1006 ý giải rõ giúp mik nha . Cảm ơn nhiều!
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A = 2011.2013
A = 2011.(2012+1)
A = 2011.2012+2011
B = 20122
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