mik biet lam nhung ngại viet lam ban thong cam nha
Tìm x, biết:
\(a)\frac{x+1}{65}+\frac{x+2}{64}=\frac{x+3}{63}+\frac{x+4}{62}\)
\(b)\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)
CMR: \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{101}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+ \frac{1}{102}\right)=\frac{1}{52}+\frac{1}{53}+...+\frac{1}{102}\)
CMR
\(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+\frac{5}{3^5}+.....+\frac{102}{3^{102}}<\frac{3}{4}\)
CMR:
a, \(100-\left(1+\frac{1}{2}+\frac{1}{3}+..+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+..+\frac{99}{100}\)
b, \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+..+\frac{1}{200}\right)=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
Giải nhanh giùm mình nhé!!!!!!!!!!!!!!
CMR \(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+\frac{5}{3^5}+....+\frac{102}{3^{102}}
CMR:
a)\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\) <1
b)\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{4^4}+....+\frac{101}{3^{101}}\),<3/4
nhanh nhé
CMR:
\(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{299}+\frac{1}{300}>\frac{2}{3}\)\(\frac{2}{3}\)
CMR \(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{102}{3^{102}}\)
Giúp mình đi mà!! Cần Gấp lắm rồi
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{200}=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)
