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2014+2015+2016/2015+2016+2017<2014/2015+2015/2016+2016/2017
B = \(\frac{2015+2016+2017}{2016+2017+2018}=\frac{2016.3}{2017.3}=\frac{2016}{2017}\left(1\right)\)
Mà A = \(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}.\left(2\right)\)
Từ \(\left(1\right)\)và \(\left(2\right)\)=> A > B.
Vậy A > B .
Bạn Dont look at me
Bạn nên làm theo bạn ấy
Bạn k đúng cho bạn ấy. Bởi vì bạn ấy làm đúng
Theo mk là vậy
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 = \(\frac{2015^{2016}+1}{2015^{2015}+1}=\frac{2015^{2015}+1}{2015^{2015}+1}+\frac{2015}{2015^{2015}+1}=1+\frac{2015}{2015^{2015}+1}\)
B = \(\frac{2014^{2015}+1}{2014^{2014}+1}=\frac{2014^{2014}+1}{2014^{2014}+1}+\frac{2014}{2014^{2014}+1}=1+\frac{2014}{2014^{2014}+1}\)
Rồi bạn tự so sánh nha
Ta có : P = 2014/2015 + 2015/2016 + 2016/2017 < 2014/(2015+2016+2017) + 2015/(2015+2016+2017) + 2016/(2015+2016+2017) = Q
Suy ra : P < Q
Vậy P < Q.
Ta thấy:\(\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2017}\)>\(\frac{2014+2015+2016}{2015+2016+2017}\)
Vậy :P>Q
Đặt A= 2015^2013+1/2015^2014+7, B=2015^2014-2/2015^2015-2
2015A= 2015^2014+2015/2015^2014+7= 1 + (2008/2015^2014+7)
2015B= 2015^2015-4030/2015^2015-2= 1 - (4028/2015^2015-2)
Do 2015A>1>2015B nên A>B
phân tích B ta có
B = \(\frac{2014+2015}{2015+2016}=\frac{2014}{2015+2016}+\frac{2015}{2015+2016}\)
vì \(\frac{2014}{2015+2016}<\frac{2014}{2015}\) ., \(\frac{2015}{2015+2016}<\frac{2015}{2016}\)
=> B< A
A=2014/2015+2015/2016. B=(2014+2015)/(2015+2016)
A=1-1/2015+1-1/2016. B=1-2/4031
A=1+1-(2015+2016)/(2015x2016). So sánh
A=1+1-(4031)/(2015x2x1008). 1+1-[4031/(4030x1008)]>1;1-2/4031<1.
A=1+1-[4031/(4030x1008)]. Vậy 1+1-[4031/(4030x1008)]>1-2/4031.
=>A>B
Đặt \(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{2015}-\frac{1}{2016}\)
\(A=\left(1+\frac{1}{3}+\frac{1}{5}+.....+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2016}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}+\frac{1}{2016}\right)-2\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2016}\right)\)
\(A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2015}+\frac{1}{2016}-\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{1008}\right)\)
\(A=\frac{1}{1009}+\frac{1}{1010}+.....+\frac{1}{2016}\)
Khi đó \(\frac{\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2015}-\frac{1}{2016}\right)}{\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}}=\frac{A}{\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}}=\frac{\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}}{\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}}=1\)
Bạn xem lời giải của mình nhé:
Giải:
Bài 2:
Ta xét A = \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(=1+\left(\frac{1}{2}-1\right)+\frac{1}{3}+\left(\frac{1}{4}-\frac{2}{4}\right)+...+\frac{1}{2015}+\left(\frac{1}{2016}-\frac{2}{2016}\right)\\ =1+\frac{1}{2}-1+\frac{1}{3}+\frac{1}{4}-\frac{1}{2}+...+\frac{1}{2015}+\frac{1}{2016}-\frac{1}{1008}\)
\(=\left(1-1\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+...+\left(\frac{1}{1008}-\frac{1}{1008}\right)+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\)
\(=\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\)
\(\Rightarrow\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\right):\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right)\\ =\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right):\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right)\\ =1\)
Chúc bạn học tốt!
Ta có: \(\frac{2015}{-2014}=-\frac{2015}{2014}=-\left(1+\frac{1}{2014}\right)\)\(=-1-\frac{1}{2014}\)
\(\frac{-2016}{2015}=-\frac{2016}{2015}=-\left(1+\frac{1}{2015}\right)\)\(=-1-\frac{1}{2015}\)
Do \(\frac{1}{2014}\) > \(\frac{1}{2015}\) => \(\frac{2015}{-2014}\) < \(\frac{-2016}{2015}\)