\(\frac{1}{n}+\frac{2015}{2014}+\frac{2014}{2013}+\frac{1}{n+1}\)
1) CMR : A=(n+2015)(n+2016) + n2 + n chia hết cho 2 với n ϵ N
2) So sánh :
P = \(\frac{2013}{2014^{2013}}+\frac{2014}{2015^{2014}}+\frac{2015}{2016^{2015}}+\frac{2016}{2017^{2016}}\) và
Q = \(\frac{2014}{2017^{2016}}+\frac{2013}{2016^{2015}}+\frac{2016}{2015^{2014}}+\frac{2015}{2014^{2013}}\)
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)
Tính:
\(\frac{1}{1+\frac{2013}{2014}+\frac{2013}{2015}}+\frac{1}{1+\frac{2014}{2015}+\frac{2014}{2013}}+\frac{1}{1+\frac{2015}{2013}+\frac{2015}{2014}}\)
tìm n thỏa mãn
\(\frac{1}{n}+\frac{2015}{2014}=\frac{2014}{2013}+\frac{1}{n+1}\)
1/n+1+1/2014=1+1/2013+1/(n+1)
1/n-1/(N+1)=1/2013-1/2014
1/n*(n+1)=1/(2013*2014)
Do do n=2013
Nho k cho mink nha
\(=>\frac{1}{n}-\frac{1}{n+1}\) \(=\frac{2014}{2013}-\frac{2015}{2014}\)
\(=>\frac{1}{n.\left(n+1\right)}\) \(=\frac{1}{4054182}\)
\(=>n.\left(n+1\right)\) \(=4054182\)
\(=>n=2013\)
Tk mk nhé
\(\frac{1}{n}+\frac{2015}{2014}=\frac{2014}{2013}+\frac{1}{n+1}\)
\(\Rightarrow\frac{1}{n}+1+\frac{1}{2014}=1+\frac{1}{2013}+\frac{1}{n+1}\)
\(\Rightarrow\frac{n+1}{n}+\frac{1}{2014}=\frac{1}{2013}+\frac{n+2}{n+1}\)
Vậy................
1. So sánh M và N ( Ko Quy Đồng)
biết M = \(\frac{2012}{2013}+\frac{2013}{2014}+\frac{2014}{2015}\)và
N =\(\frac{2012+2013+2014}{2013+2014+2015}\)
( Giải rõ ràn nha) tớ tick cho
\(N=\frac{2012+2013+2014}{2013+2014+2015}=\frac{2012}{2013+2014+2015}+\frac{2013}{2013+2014+2015}+\frac{2014}{2013+2014+2015}\)
Ta thấy: \(\frac{2012}{2013}>\frac{2012}{2013+2014+2015}\)
\(\frac{2013}{2014}>\frac{2013}{2013+2014+2015}\)
\(\frac{2014}{2015}>\frac{2014}{2013+2014+2015}\)
\(\Rightarrow M=\frac{2012}{2013}+\frac{2013}{2014}+\frac{2014}{2015}>N=\frac{2012}{2013+2014+2015}+\frac{2013}{2013+2014+2015}+\frac{2014}{2013+2014+2015}\)
Vậy M>N
A=\(\frac{2014+\frac{2013}{2}+\frac{2012}{3}+.....+\frac{2}{2013}+\frac{1}{2014}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2014}+\frac{1}{2015}}\)=
Xét Tử số của A ta có:
\(2014+\frac{2013}{2}+\frac{2012}{3}+....+\frac{2}{2013}=1+\left(\frac{2013}{2}+1\right)+\left(\frac{2012}{3}+1\right)+....+\left(\frac{1}{2014}+1\right)\)\(TS=\frac{2015}{2}+\frac{2015}{3}+....+\frac{2015}{2014}+\frac{2015}{2015}\)
\(TS=2015.\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2015}\right)\)
\(A=\frac{2015.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)}{\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2015}\right)}=2015\)
toán lớp 8 dễ quá vậy
A=2015
hình như thế
giá trị của biểu thức A=\(\frac{2014+\frac{2013}{2}+\frac{2012}{3}+....+\frac{2}{2013}+\frac{1}{2014}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}+\frac{1}{2015}}\)
a, Cho A=\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+...+\frac{1}{99}+\frac{1}{100}\) . So Sánh A với 1
b, B=\(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}\). So sánh B với \(\frac{1}{2}\)
c, cho M=\(\frac{2013}{2014}+\frac{2014}{2015}\)và N=\(\frac{2013+2014}{2014+2015}\). So sánh M và N
Câu a, p/s cuối cùng là \(\frac{1}{100}\)nha mí bn
a) Ta có :
\(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{100}\)
\(>\frac{1}{10}+\frac{1}{100}.90=\frac{1}{10}+\frac{90}{100}=1\)
vậy A > 1
b) \(B=\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}\)
\(>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=\frac{1}{20}.10=\frac{1}{2}\)
Vậy B > \(\frac{1}{2}\)
giá trị biểu thức A=\(\frac{2014+\frac{2013}{2}+\frac{2012}{3}+...+\frac{2}{2013}+\frac{1}{2014}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}+\frac{1}{2015}}là?\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}+\frac{1}{2016}}{\frac{2015}{1}+\frac{2014}{2}+\frac{2013}{3}+...+\frac{1}{2014}+\frac{1}{2015}}\)
xét mẫu(chỗ 1/2014 sửa lại thành 2/2014)
=(1/2015+1)+(2/2014+1)+...+(2013/3+1)+(2014/2+1)+(2015/1-2014)
=2016/2015+2016/2014+...+2016/3+2016/2+1
=2016.(1/2016+1/2015+...+1/4+1/3+1/2)
=> A= 1/2016
mún dễ hỉu hơn hãy gửi tin nhắn cho mik