CMR: Nếu \(\frac{a+2014}{a-2014}=\frac{b+2015}{b-2015}thì\frac{a}{2014}=\frac{b}{2015}\)
Chứng minh rằng nếu có: \(\frac{a+2014}{a-2014}=\frac{b+2015}{b-2015}\)thì: \(\frac{a}{2014}=\frac{b}{2015}\)
CMR : \(\frac{a+2014}{a-2014}\)= \(\frac{b+2015}{b-2015}\) thì \(\frac{a}{2014}=\frac{b}{2015}\) Với a,b thuộc Z
\(\frac{a+2014}{a-2014}=\frac{b+2015}{b-2015}\Rightarrow\left(a+2014\right)\left(b-2015\right)=\left(a-2014\right)\left(b+2015\right)\)
\(\Rightarrow\) \(ab+2014b-2015a-2014.2015=ab+2015a-2014b-2014.2015\)
\(\Rightarrow\) \(\left(ab-ab\right)+\left(-2014.2015+2014.2015\right)=\left(2015a+2015a\right)-\left(2014b+2014b\right)\)
\(\Rightarrow0+0=4030a-4028b\)
\(\Rightarrow4030a=4028b\) \(\Rightarrow\frac{a}{b}=\frac{4028}{4030}=\frac{2014}{2015}\Rightarrow\frac{a}{2014}=\frac{b}{2015}\)
Vậy nếu \(\frac{a+2014}{a-2014}=\frac{b+2015}{b-2015}\) thì \(\frac{a}{2014}=\frac{b}{2015}\) (đpcm)
Ta có : \(\frac{a+2014}{a-2014}=\frac{a+2015}{a-2015}\)
\(\Rightarrow\left(a+2014\right)\left(a-2015\right)=\left(a-2014\right)\left(a+2015\right)\)
\(\Rightarrow a^2-a-2014.2015=a^2+a-2014.2015\)
\(\Leftrightarrow a^2-a=a^2+a\)
=> a2 - a2 - a = a
=> -a = a
=> 0 = a + a
=> 2a = 0
=> a = 0
Vậy \(\frac{a}{2014}=\frac{b}{2015}\) (đpcm)
cho a>b> 0 chưng minh : \(\frac{a^{2015}-b^{2015}}{a^{2015}+b^{2015}}>\frac{a^{2014}-b^{2014}}{a^{2014}+b^{2014}}\)
Chia cả tử và mẫu của mỗi phân số tương ứng cho b2015; b2014
=> cần chứng minh: \(\frac{\left(\frac{a}{b}\right)^{2015}-1}{\left(\frac{a}{b}\right)^{2015}+1}>\frac{\left(\frac{a}{b}\right)^{2014}-1}{\left(\frac{a}{b}\right)^{2014}+1}\)
Ta có: \(VT=\frac{\left(\frac{a}{b}\right)^{2015}-1}{\left(\frac{a}{b}\right)^{2015}+1}=\frac{\left(\frac{a}{b}\right)^{2015}+1}{\left(\frac{a}{b}\right)^{2015}+1}-\frac{2}{\left(\frac{a}{b}\right)^{2015}+1}=1-\frac{2}{\left(\frac{a}{b}\right)^{2015}+1}\)
\(VP=\frac{\left(\frac{a}{b}\right)^{2014}-1}{\left(\frac{a}{b}\right)^{2014}+1}=\frac{\left(\frac{a}{b}\right)^{2014}+1}{\left(\frac{a}{b}\right)^{2014}+1}-\frac{2}{\left(\frac{a}{b}\right)^{2014}+1}=1-\frac{2}{\left(\frac{a}{b}\right)^{2014}+1}\)
Vì a> b > 0 => a/b > 1. Do đó:
\(\left(\frac{a}{b}\right)^{2015}+1>\left(\frac{a}{b}\right)^{2014}+1\)
=> \(\frac{2}{\left(\frac{a}{b}\right)^{2015}+1}1-\frac{2}{\left(\frac{a}{b}\right)^{2014}+1}\)
=> VT > VP
So sánh 2 số:
\(a)\sqrt{2014}-\sqrt{2013};B=\sqrt{2015}-\sqrt{2014}\\ b)E=\frac{2014}{\sqrt{2015}}+\frac{2015}{\sqrt{2014}};F=\sqrt{2014}+\sqrt{2015}\)
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 : \(A=\frac{2015^{2016}+1}{2015^{2015}+1}\) và \(B=\frac{2014^{2015}+1}{2014^{2014}+1}\)
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
So sánh
A=\(\frac{2015^{2014}+1}{2015^{2014}-1}\) B=\(\frac{2015^{2014}-1}{2015^{2014}-3}\)
CÁCH 1:
A=1và 2/2015^2014-1
B= 1và 2/2015^2014-3
Vì 1và 2/2015^2014-1 < 1và 2/2015^2014-3
Vậy A <B
CÁCH 2:
Ta biết: a/b>1=>a/b> a+n/b+n
B>1=> B= 2015^2014-1/2015^2014-3> 2015^2014-1+2/2015^2014-3+2=2015^2014+1/2015^2014-1=A
Vậy B>A
Cho:
A = \(\frac{2015}{2014^2+1}\)+\(\frac{2015}{2014^2+2}\)+\(\frac{2015}{2014^2+2014}\)
CMR: A không thuộc N !?!?!?