Cho M=\(\frac{2013}{2014}+\frac{2014}{2015}\)
N=\(\frac{2013+2014}{2014+2015}\)
So sánh M và N
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
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)
So sánh M và N
M = \(\frac{2012}{2013}\)+ \(\frac{2013}{2014}\)+ \(\frac{2014}{2015}\)
N = \(\frac{2012+2013+2014}{2013+2014+2015}\)
Giúp mình với
Xét N có:
\(N=\frac{2012+2013+2014}{2013+2014+2015}=\frac{2012}{2013+2014+2015}+\frac{2013}{2013+2014+2015}+\frac{2014}{2013+2014+2015}\)
Ta các số hạng của M và N có:
\(\frac{2012}{2013}>\frac{2012}{2013+2014+2015}\) (1)
\(\frac{2013}{2014}>\frac{2013}{2013+2014+2015}\) (2)
\(\frac{2014}{2015}>\frac{2014}{2013+2014+2015}\) (3)
Từ (1);(2);(3) => M > N
a) So sánh \(\frac{2013}{2015}\) và \(\frac{2014}{2016}\)
b) So sánh \(\frac{2013+2014}{2014+2015}\) và \(\frac{2013}{2014}+\frac{2014}{2015}\)
a)\(\frac{2013}{2015}< \frac{2014}{2016}\)
b)\(\frac{2013+2014}{2014+2015}< \frac{2013}{2014}+\frac{2014}{2015}\)
ta có tính chất \(\frac{a}{b}\)>1 suy ra \(\frac{a.m}{b.m}\).........
so sánh
\(\frac{2013}{2014}+\frac{2014}{2015}và\frac{2013+2014}{2014+2015}\)
Ta có: \(\frac{2013}{2014}>\frac{2013}{2014+2015}\) (1)
\(\frac{2014}{2015}>\frac{2014}{2014+2015}\) (2)
ộng caác bất đẳng thứa (1) và (2) vào vế với vế:
\(\frac{2013}{2014}+\frac{2014}{2015}>\frac{2013+2014}{2014+2015}\Rightarrow A>B\)
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So sánh:
\(\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}\)và\(\frac{2013+2014+2015}{2014+2015+2016}\)
so sánh\(\frac{n-2013}{n-2014}và\frac{n-2014}{n-2015}\)
Ta có :
\(\frac{n-2013}{n-2014}=1-\frac{2013}{2014}=\frac{1}{2014}\)
\(\frac{n-2014}{n-2015}=1-\frac{2014}{2015}=\frac{1}{2015}\)
Vì \(\frac{1}{2014}>\frac{1}{2015}\Rightarrow\frac{n-2013}{n-2014}<\frac{n-2014}{n-2015}\)
\(\frac{n-2013}{n-2014}<\frac{n-2014}{n-2015}\)
So sánh C= \(\frac{2013}{2014}+\frac{2014}{2015}\)
và D=\(\frac{2013+2014}{2014+2015}\)
D\(\frac{2013}{2014+2015}+\frac{2014}{2014+2015}\)
Vì \(\frac{2013}{2014}>\frac{2013}{204+2015}\)
và \(\frac{2014}{2015}>\frac{2014}{2014+2015}\)
nên C>D
Ủng hộ mk nha
\(\frac{2013}{2014}+\frac{2014}{2015}=1,999...\)
\(\frac{2013+2014}{2014+2015}=4029\)
nen D>C
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 > \(\frac{1}{2}\)