tìm n thỏa mãn
\(\frac{1}{n}+\frac{2015}{2014}=\frac{2014}{2013}+\frac{1}{n+1}\)
Những câu hỏi liên quan
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)
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\(\frac{1}{n}+\frac{2015}{2014}+\frac{2014}{2013}+\frac{1}{n+1}\)
bài 1)trung bình cộng các giá trị x thỏa mãn 4(x-1)^2=x^2
bài 2)\(\frac{2014+\frac{2013}{2}+\frac{2012}{3}+.............+\frac{2}{2013}+\frac{1}{2014}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..............+\frac{1}{2014}+\frac{1}{2015}}\)
2) xét tử ta có
2014+2013/2+2012/3+...+2/2013+1/2014
=(1+2013/2)+(1+2012/3)+...+(1+2/2013)+(1+1/2014)+1
=2015/2+2015/3+...+2015/2013+2015/2014+2015/2015
=2015(1/2+1/3+...+1/2013+1/2014+1/2015) (1)
mà mẫu bằng 1/2+1/3+1/4+...+1/2014+1/2015 (2)
từ (1),(2)=> phân thức trên =2015
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Tính:
\(\frac{1}{1+\frac{2013}{2014}+\frac{2013}{2015}}+\frac{1}{1+\frac{2014}{2015}+\frac{2014}{2013}}+\frac{1}{1+\frac{2015}{2013}+\frac{2015}{2014}}\)
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
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a, Cho Afrac{1}{10}+frac{1}{11}+frac{1}{12}+frac{1}{13}+frac{1}{14}+...+frac{1}{99}+frac{1}{100} . So Sánh A với 1b, Bfrac{1}{11}+frac{1}{12}+...+frac{1}{20}. So sánh B với frac{1}{2}c, cho Mfrac{2013}{2014}+frac{2014}{2015}và Nfrac{2013+2014}{2014+2015}. So sánh M và NCâu a, p/s cuối cùng là frac{1}{100}nha mí bn
Đọc tiếp
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 > \(\frac{1}{2}\)
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Xem thêm câu trả lời
Tìm x thỏa mãn:
\(\frac{x+1}{2018}+\frac{x+2}{2017}+\frac{x+3}{2016}=\frac{x+4}{2015}+\frac{x+5}{2014}+\frac{x+6}{2013}\)
\(\frac{x+1}{2018}+\frac{x+2}{2017}+\frac{x+3}{2016}=\frac{x+4}{2015}+\frac{x+5}{2014}+\frac{x+6}{2013}\)
\(\Leftrightarrow\) \(\frac{x+1}{2018}+1+\frac{x+2}{2017}+1+\frac{x+3}{2016}+1=\frac{x+4}{2015}+1+\frac{x+5}{2014}+1+\frac{x+6}{2013}+1\)
\(\Leftrightarrow\frac{x+2019}{2018}+\frac{x+2019}{2017}+\frac{x+2019}{2016}=\frac{x+2019}{2015}+\frac{x+2019}{2014}+\frac{x+2019}{2013}\)
\(\Leftrightarrow\frac{x+2019}{2018}+\frac{x+2019}{2017}+\frac{x+2019}{2016}-\frac{x+2019}{2015}-\frac{x+2019}{2014}-\frac{x+2019}{2013}=0\)
\(\Leftrightarrow\left(x+2019\right)\left(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}-\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\right)\)\(=0\)
Lại có: \(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}-\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\) \(\ne\) \(0\)
\(\Rightarrow x+2019=0\)
\(\Rightarrow x=0-2019=-2019\)
Vậy x= -2019
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Tìm x biết
\(\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2014}+\frac{1}{2015}\right).x=\frac{2014}{1}+\frac{2013}{2}+\frac{2012}{3}+...+\frac{2}{2013}+\frac{1}{2014}\)
có 2014/1+2013/2+2012/3+...+2/2013+1/2014=[1+(2013/2)]+[1+(2012/3)]+...+[1+(2/2013)]+[1+(1/2014)]+1
=2015/2+2015/3+...+2015/2014+2015/2015=2015.[1/2+1/3+..+1/2015)
vậy (1/2+1/3+...+1/2015).x=(1/2+1/3+...+1/2015).2015
x=2015
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Cho M=\(\frac{2013}{2014}+\frac{2014}{2015}\)
N=\(\frac{2013+2014}{2014+2015}\)
So sánh M và N
Ta có:
\(\frac{2013}{2014}>\frac{2013}{2014+2015}\)
\(\frac{2014}{2015}>\frac{2014}{2014+2015}\)
\(\Rightarrow\frac{2013}{2014}+\frac{2014}{2015}>\frac{2013+2014}{2014+2015}\)
\(\Rightarrow M>N\)
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Ta có: \(N=\frac{2013+2014}{2014+2015}<1\);
\(M=\frac{2013}{2014}+\frac{2014}{2015}>\frac{2013}{2015}+\frac{2014}{2015}=\frac{4027}{2015}>1\)
\(\Rightarrow A>B\)
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TA CÓ
2013>\(\frac{2013}{2014+2015}\)
2014>\(\frac{2014}{2014+2015}\)
=>2013+2014/2014+ 2015>2013+2014/2014+2015
=>M>N
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