Các câu hỏi tương tự
Tìm x,y \(\in Z\):
|x-3|.|x+3|=16
Chứng minh:
\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{2016^2}< \frac{1}{2}\)
So sánh:
\(A=\frac{1999^{1999}+1}{1999^{2000}+1}\)và \(B=\frac{1999^{1998}+1}{1999^{1999}+1}\)
So sánh:
a) M=\(\frac{1999^{1999+1}}{1999^{2000}+1}và\)N=\(\frac{1999^{1989}+1}{1999^{2009}+1}\)
b) A=\(\frac{-9}{10^{2010}}+\frac{-19}{10^{2011}}và\)B=\(\frac{-9}{10^{2011}}+\frac{-19}{10^{2010}}\)
So sánh
\(C=\frac{1999^{2000}+1}{1999^{1999}+1}\)và \(D=\frac{1999^{1999}+1}{1999^{1998}+1}\)
tính hợp lí
\(\frac{1999\cdot2010-1999}{2010\cdot1998-2010}\)
\(\frac{2}{35}+\frac{4}{77}+\frac{2}{143}+\frac{4}{221}+\frac{2}{323}+\frac{4}{437}+\frac{2}{575}\)
bài trên có sai không ạ
\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{20}}{\frac{1999}{1}+\frac{1998}{2}+\frac{1997}{3}+...+\frac{1}{1999}}\)
\(E=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}{\frac{1999}{1}+\frac{1998}{2}+\frac{1997}{3}+..+\frac{1}{1999}}\)
Tính
A=\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2000}}{\frac{1999}{1}+\frac{1998}{2}+\frac{1997}{3}+......+\frac{1}{1999}}\)
Ai nhanh và đúng mình tick cho
cho A=\(\frac{79}{1999}+\frac{191}{1998}+\frac{947}{1997}+\frac{673}{1998}+\frac{110}{1999}\)
so sánh A với 1
Tính \(E=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}{\frac{1999}{1}+\frac{1998}{2}+\frac{1997}{3}+...+\frac{1}{1999}}\)