\(Cho\) \(E=\frac{100^2+1^2}{100.1}+\frac{99^2+2^2}{99.2}+\frac{98^2+3^2}{98.3}+...+\frac{52^2+49^2}{52.49}+\frac{51^2+50^2}{51.50}\)
\(Cho\) \(F=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}+\frac{1}{101}\) \(;\) \(Cho\)\(G=\frac{1}{100.1}+\frac{1}{99.2}+\frac{1}{98.3}+...+\frac{1}{52.49}+\frac{1}{51.50}\)
\(a,Tính:\) \(\frac{E}{F}\) \(b,Tính:\) \(F-101G\)
\(Làm\) \(giúp\)\(mình\)\(nha!\)\(Đang\)\(cần\)\(!\)
Cho \(E=\frac{100^2+1^1}{100.1}+\frac{99^2+2^2}{99.2}+\frac{98^2+3^2}{98.3}+...+\frac{52^2+49^2}{52.49}+\frac{51^2+50^2}{51.50}\)
\(F=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...\frac{1}{101}\)
\(G=\frac{1}{100.1}+\frac{1}{99.2}+...\frac{1}{51.50}\)
a) Tính \(\frac{E}{F}\)
b) Tính F - 101G
Các bạn giúp mình nhé!
\(ChoE=\frac{100^2+1^2}{100.1}+\frac{99^2+2^2}{99.2}+\frac{98^2+3^2}{98.3}+...+\frac{52^2+49^2}{52.49}+\frac{51^2+50^2}{51.50}\)
\(ChoF=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}+\frac{1}{101}\)
\(ChoG=\frac{1}{100.1}+\frac{1}{99.2}+\frac{1}{98.3}+...+\frac{1}{52.49}+\frac{1}{51.50}\)
\(a.Tính:\frac{E}{F}\) \(b.Tính:F-101G\)
\(ChoE=\frac{100^2+1^2}{100.1}+\frac{99^2+2^2}{99.2}+\frac{98^2+3^2}{98.3}+...+\frac{52^2+49^2}{52.49}+\frac{51^2+50^2}{51.50}\)
\(ChoF=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}+\frac{1}{101}\)
\(ChoG=\frac{1}{100.1}+\frac{1}{99.2}+\frac{1}{98.3}+...+\frac{1}{52.49}+\frac{1}{51.50}\)
\(a.Tính:\frac{E}{F}\) \(b.Tính:F-101G\)
\(ChoE=\frac{100^2+1^2}{100.1}+\frac{99^2+2^2}{99.2}+\frac{98^2+3^2}{98.3}+...+\frac{52^2+49^2}{52.49}+\frac{51^2+50^2}{51.50}\)
\(ChoF=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}+\frac{1}{101}\)
\(ChoG=\frac{1}{100.1}+\frac{1}{99.2}+\frac{1}{98.3}+...+\frac{1}{52.49}+\frac{1}{51.50}\)
\(a.Tính:\frac{E}{F}\) \(b.Tính:F-101G\)
Cho \(E=\frac{100^2+1^2}{100.1}+\frac{99^2+2^2}{99.2}+\frac{98^2+3^2}{98.3}+...+\frac{52^2+49^2}{52.49}+\frac{51^2+50^2}{51.50}\)
F = 1/2+1/3+1/4+1/5+...+1/100+1/101
Tính E/F
\(ChoE=\frac{100^2+1^2}{100.1}+\frac{99^2+2^2}{99.2}+\frac{98^2+3^2}{98.3}+...+\frac{52^2+49^2}{52.49}+\frac{51^2+50^2}{51.50}\)
\(ChoF=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}+\frac{1}{101}\)
\(ChoG=\frac{1}{100.1}+\frac{1}{99.2}+\frac{1}{98.3}+...+\frac{1}{52.49}+\frac{1}{51.50}\)
\(a.Tính:\frac{E}{F}\) \(b.Tính:F-101G\)
cho E = \(\frac{100^2+1^2}{100.1}+\frac{99^2+2^2}{99.2}+\frac{98^2+3^2}{98.3}+...+\frac{52^2+49^2}{52.49}+\frac{50^2+49^2}{50.49}\)
F = \(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}+\frac{1}{101}\)
Hãy tính E/F
\(Cho\) \(E=\frac{100^2+1^2}{100.1}+\frac{99^2+2^2}{99.2}+\frac{98^2+3^2}{98.3}+...+\frac{52^2+49^2}{52.49}+\frac{51^2+50^2}{51.50}\)
\(Cho\) \(F=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}+\frac{1}{101}\) \(;\) \(Cho\)\(G=\frac{1}{100.1}+\frac{1}{99.2}+\frac{1}{98.3}+...+\frac{1}{52.49}+\frac{1}{51.50}\)
\(a,Tính:\) \(\frac{E}{F}\) \(b,Tính:\) \(F-101G\)
\(Làm\) \(giúp\)\(mình\)\(nha!\)\(Đang\)\(cần\)\(!\)
\(A=\frac{100^2+1^2}{100.1}+\frac{99^2+2^2}{99.2}+...+\frac{52^2+49^2}{52.49}+\frac{51^2+50^2}{51.50}\)
\(B=\frac{1}{100.1}+\frac{1}{99.2}+...+\frac{1}{51.50}\)
\(C=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}+\frac{1}{101}\)
a) Tính\(\frac{A}{C}\)
b)Tính C-101B
