So sánh 2 số A=2013.2015 và B=21142
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DQ
5
NC
1
HT
8 tháng 9 2016
A = 2012.2016 = (2013 - 1)(2015 + 1)
= 2013.2015 + 2013 - 2015 - 1
= 2013.2015 - 3 < 2013.2015
=> A < B
NT
2
KN
15 tháng 7 2019
Dễ c/m đẳng thức: \(\left(n-1\right)\left(n+1\right)=n^2-1\)
Lúc đó: \(A=2014^2-1+2015^2-1=2014^2+2015^2-2=B\)
Vậy A = B
CC
15 tháng 7 2019
\(A=2013.2015+2014.2016\)
\(=\left(2015-2\right).2015+2014\left(2014+2\right)\)
\(=(2015^2-4030)+(2014^2+4028)\)
\(=\left(2015^2+2014^2\right)-\left(4030-4028\right)\)
\(=2014^2+2015^2-2\)
\(\Rightarrow A=B\)
NT
1
CT
0
6 tháng 10 2018
1)
a)\(A=2013.2015=2013.\left(2014+1\right)=2013.2014+2013\)
\(B=2014^2=2014.\left(2013+1\right)=2014.2013+2014\)
Ta có: \(2014.2013+2014>2013.2014+2013\)
\(\Rightarrow2014^2>2013.2015\)
\(\Rightarrow B>A\)
Vậy \(B>A\)
b) \(A=4.\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)
\(\Rightarrow2A=2.4\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3-1\right).\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^8-1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^{16}-1\right)...\left(3^{64}+1\right)\)
\(\Rightarrow2A=3^{128}-1\)
\(\Rightarrow A=\frac{3^{128}-1}{2}< 3^{128}-1=B\)
\(\Rightarrow A< B\)
Vậy \(A< B\)
2)
a)\(9x^2-6x+3=\left(3x\right)^2-2.3x.1+1^2+2\)
\(=\left(3x-1\right)^2+2\)
Ta có: \(\left(3x-1\right)^2\ge0\forall x\)
\(\Rightarrow\left(3x-1\right)^2+2\ge2\forall x\)
\(\Rightarrow\left(3x-1\right)^2+2>0\forall x\)
đpcm
b)\(x^2+y^2+2x+6y+16\)
\(=\left(x^2+2x+1\right)+\left(y^2+2.y.3+3^2\right)+6\)
\(=\left(x+1\right)^2+\left(y+3\right)^2+6\)
Ta có: \(\hept{\begin{cases}\left(x+1\right)^2\ge0\forall x\\\left(y+3\right)^2\ge0\forall y\end{cases}\Rightarrow}\left(x+1\right)^2+\left(y+3\right)^2+6\ge6\forall x;y\)
\(\Rightarrow\left(x+1\right)^2+\left(y+3\right)^2+6>0\)
đpcm
Tham khảo nhé~
6 tháng 10 2018
1.
a) A = 2013.2015 = (2014 - 1)(2014 + 1) = 20142 - 1
Vì 20142 - 1 < 20142 => A < B
Vậy A < B
b) \(A=4\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^8-1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)
\(\Rightarrow2A=3^{128}-1\Leftrightarrow A=\frac{3^{128}-1}{2}\)
\(\Rightarrow A< B\)
Vậy A < B
Bài 2:
a) \(9x^2-6x+2=\left(3x\right)^2-2.3x+1+2=\left(3x-1\right)^2+2\)
Vì \(\left(3x-1\right)^2\ge0\Rightarrow\left(3x-1\right)^2+2>0\)
=> 9x2 - 6x + 2 luôn nhận giá trị dương với mọi x
b) \(x^2+y^2+2x+6y+16=\left(x^2+2x+1\right)+\left(y^2+6y+9\right)+6=\left(x+1\right)^2+\left(y+3\right)^2+6\)
Vì \(\left(x+1\right)^2\ge0;\left(y+3\right)^2\ge0\Rightarrow\left(x+1\right)^2+\left(y+3\right)^2\ge0\Rightarrow\left(x+1\right)^2+\left(y+3\right)^2+6>0\)
=> x2 + y2 + 2x + 6y + 16 luôn nhận giá trị dương với mọi x
21 tháng 2 2022
a) ta có: \(1-\frac{2012}{2013}=\frac{1}{2013}\)
\(1-\frac{2013}{2014}=\frac{1}{2014}\)
mà \(\frac{1}{2013}>\frac{1}{2014}\) nên \(\frac{2013}{2014}>\frac{2012}{2013}\)
\(A=2013.2015