Vì \(\frac{n}{m+2017}=\frac{2017}{m+n}\Rightarrow n\left(m+n\right)=2017\left(m+2017\right)\Rightarrow n=2017\)
\(\frac{m}{n+2017}=\frac{2017}{m+n}\Rightarrow2017\left(n+2017\right)=m\left(m+n\right)\Rightarrow m=2017\)
\(\Rightarrow x=\frac{2017}{2017+2017}=\frac{2017}{2017+2017}=\frac{2017}{2017+2017}=\frac{1}{2}\)
\(\frac{m}{n}\)=\(\frac{2017}{2017}\)chứng minh rằng \(\frac{m}{n}\)=\(\frac{m+2017}{n+2017}\)
So sánh M và N biết:
M=\(\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2017}\)
N=\(\frac{2014+2015+2016}{2015+2016+2017}\)
So sánh \(A=\frac{2016}{a^m}+\frac{2016}{a^n}vaB=\frac{2017}{a^m}\frac{2015}{a^n}\)
Cho tổng A=\(\frac{2018}{2017^2+1}+\frac{2018}{2017^2+2}+\frac{2018}{2017^2+3}+...+\frac{2018}{2017^2+n}+...+\frac{2018}{2017^2+2017}\)
(A có 2017 số hạng). Chứng tỏ A không là số nguyên
1.cho Q(x)=(m-1).x^3+(m-n).x^2+(n-1).x+2(m-n)
Tim m,n để Q(x)=0 với mọi x
2.Tim x,y thoả mãn
|x+2018|^2019+|y-2017|^2017
So sánh
\(A=\frac{2018^n-2017^n}{2018^n+2017^n}\)+\(\frac{2017^n-2016^n}{2017^n+2016^n}\)
Tính \(\left(M-N\right)^2\)với \(M=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}\)
và \(N=\frac{1}{1010}+\frac{1}{1011}+....+\frac{1}{2019}\)
Cho a,b,c là các số thực khác 0 thỏa mãn : \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}.\)
Tìm a,b,c biết rằng: 3a2017+4b2017+5c2017=12x52017
Cho \(S=\frac{2}{2017+1}+\frac{2^2}{2017^2+1}+...+\frac{2^{n+1}}{2017^{2^n}+1}+...+\frac{2^{2017}}{2017^{2^{2016}}+1}\)
So sanh \(S\)voi \(\frac{1}{1008}\)
