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A = (n + 2015)(n + 2016) + n2 + n
= (n + 2015)(n + 2015 + 1) + n(n + 1)
Tích 2 số tự nhiên liên tiếp luôn chia hết cho 2
=> (n + 2015)(n + 2015 + 1) chia hết cho 2
n(n + 1) chia hết cho 2
=> (n + 2015)(n + 2015 + 1) + n(n + 1) chia hết cho 2
=> A chia hết cho 2 với mọi n \(\in\) N (đpcm)
\(A=\left[1+\left(-2\right)\right]+\left[3+\left(-4\right)\right]+....+\left[2013+\left(-2014\right)+2015\right]\)
\(A=\left(-1\right)+\left(-1\right)+....+\left(-1\right)+2015\left(\text{1007 số hạng }\left(-1\right)\right)=1008\)
A = \(\frac{2013}{2014}+\frac{2014}{2015}>\frac{1}{2}+\frac{1}{2}=1\)
\(B=\frac{2013+2014+2015}{2014+2015+2016}<1\)
\(Vậy:A>B\)
Đúng nha Nguyễn Bình Minh
so sánh:
\(A=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}\) và\(B=\) \(\frac{2013+2014+2015}{2014+2015+2016}\)
\(B=\frac{2013}{2014+2015+2016}+\frac{2014}{2014+2015+2016}+\frac{2015}{2014+2015+2016}\)
Ta có: \(\frac{2013}{2014}>\frac{2013}{2014+2015+2016}\)
\(\frac{2014}{2015}>\frac{2014}{2014+2015+2016}\)
\(\frac{2015}{2016}>\frac{2015}{2014+2015+2016}\)
\(\Rightarrow\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}>\frac{2013+2014+2015}{2014+2015+2016}\)
Vậy: \(A>B\)
\(\frac{2012+2013.2014}{2014.2015-2016}=\frac{2012+2013.2014}{2014.\left(2013+2\right)-2016}\)
\(=\frac{2012+22013.2014}{2014.2013+2014.2-2016}\)
\(=\frac{2012+2013.2014}{2014.2013+2028-2016}\)
\(=\frac{2012+2013.2014}{2014.2013+2012}=1\)
\(\frac{2012+2013.2014}{2014.2015-2016}\)
\(=\frac{2012+2013.2014}{2014.\left(2013+2\right)-2016}\)
\(=\frac{2012+2013.2014}{2014.2013+2014.2-2016}\)
\(=\frac{2012}{2014.2-2016}\)
\(=\frac{2012}{4028-2016}\)
\(=\frac{2012}{2012}\)
\(=1\)
Ta có P(x)= x4+ax3+bx2+cx+d
Đặt P(x)= (x-2013)(x-2014)(x-2015)(x-x0)+mx2+nx+p
P(2013)=2014=>4052169m+2013n+p=2014} m=0
P(2014)=2015=>4056196m+2014n+p=2015}=> n=1
P(2015)=2016=>4060225m+2015n+p=2016} p=1
=>P(x)= (x-2013)(x-2014)(x-2015)(x-x0)+x+1
=>.) P(2012)= -6(2012-x0)+2012+1
= -12072+6x0+2013=-10059+6x0
.)P(2016)=6(2016-x0)+2016+1
=12096-6x0+2017=14113-6x0
=> P(2012)+P(2016)= -10059+6x0+14113-6x0=4054