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
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\(ChoE=\frac{100^2+1^2}{100.1}+\frac{99^2+2^2}{99.2}+\frac{98^2+3^2}{98.3}+...+\frac{52^2+49^2}{52.49}+\frac{51^2+50^2}{51.50}\)
\(ChoF=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}+\frac{1}{101}\)
\(ChoG=\frac{1}{100.1}+\frac{1}{99.2}+\frac{1}{98.3}+...+\frac{1}{52.49}+\frac{1}{51.50}\)
\(a.Tính:\frac{E}{F}\) \(b.Tính:F-101G\)
\(ChoE=\frac{100^2+1^2}{100.1}+\frac{99^2+2^2}{99.2}+\frac{98^2+3^2}{98.3}+...+\frac{52^2+49^2}{52.49}+\frac{51^2+50^2}{51.50}\)
\(ChoF=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}+\frac{1}{101}\)
\(ChoG=\frac{1}{100.1}+\frac{1}{99.2}+\frac{1}{98.3}+...+\frac{1}{52.49}+\frac{1}{51.50}\)
\(a.Tính:\frac{E}{F}\) \(b.Tính:F-101G\)
\(ChoE=\frac{100^2+1^2}{100.1}+\frac{99^2+2^2}{99.2}+\frac{98^2+3^2}{98.3}+...+\frac{52^2+49^2}{52.49}+\frac{51^2+50^2}{51.50}\)
\(ChoF=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}+\frac{1}{101}\)
\(ChoG=\frac{1}{100.1}+\frac{1}{99.2}+\frac{1}{98.3}+...+\frac{1}{52.49}+\frac{1}{51.50}\)
\(a.Tính:\frac{E}{F}\) \(b.Tính:F-101G\)
\(ChoE=\frac{100^2+1^2}{100.1}+\frac{99^2+2^2}{99.2}+\frac{98^2+3^2}{98.3}+...+\frac{52^2+49^2}{52.49}+\frac{51^2+50^2}{51.50}\)
\(ChoF=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}+\frac{1}{101}\)
\(ChoG=\frac{1}{100.1}+\frac{1}{99.2}+\frac{1}{98.3}+...+\frac{1}{52.49}+\frac{1}{51.50}\)
\(a.Tính:\frac{E}{F}\) \(b.Tính:F-101G\)
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
\(E=\frac{100^2+1^2}{100.1}+\frac{99^2+2^2}{99.2}+...+\frac{51^2+50^2}{51.50}\)
Tính E
\(E=\frac{100}{1}+\frac{1}{100}+\frac{99}{2}+\frac{2}{99}+...+\frac{51}{50}+\frac{50}{51}\)
\(E=\left(\frac{99}{2}+1\right)+\left(\frac{98}{3}+1\right)+...+\left(\frac{50}{51}+1\right)+1\)
\(E=\frac{101}{2}+\frac{101}{3}+...+\frac{101}{101}\)
\(E=101.\left(\frac{1}{2}+\frac{1}{3}+..+\frac{1}{101}\right)\)
Đến đây chắc tự hiểu
\(ChoE=\frac{100^2+1^2}{100.1}+\frac{99^2+2^2}{99.2}+\frac{98^2+3^2}{98.3}+...+\frac{52^2+49^2}{52.49}+\frac{51^2+50^2}{51.50}\)
\(ChoF=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}+\frac{1}{101}\)
\(ChoG=\frac{1}{100.1}+\frac{1}{99.2}+\frac{1}{98.3}+...+\frac{1}{52.49}+\frac{1}{51.50}\)
\(a.Tính:\frac{E}{F}\) \(b.Tính:F-101G\)
Cho
C= \(\frac{100^2+1^2}{100.1}+\frac{99^2+2^2}{99.2}+..................+\frac{51^2+50^2}{51.50}\)
D= \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+................+\frac{1}{1001}\)
Tính C:D
Mk sửa lại đề xíu, có lẽ bn chép sai ở phân số cuối của D phải là 1/101
C = 1002+12/100.1 + 992+22/99.2 + ... + 512+502/51.50
C = 1002/100.1 + 12/100.1 + 992/99.2 + 22/99.2 + ... + 512/51.50 + 502/51.50
C = 100/1 + 1/100 + 99/2 + 2/99 + ... + 51/50 + 50/51
C = 100/1 + 99/2 + 98/3 + ... + 51/50 + 50/51 + ... + 1/100
C = (1 + 1 + ... + 1) + 99/2 + 98/3 + ... + 1/100
100 số 1
C = (99/2 + 1) + (98/3 + 1) + ... + (1/100 + 1) + 1
C = 101/2 + 101/3 + ... + 101/100 + 101/101
C = 101.(1/2 + 1/3 + ... + 1/100 + 1/101)
=> C : D = 101
Mk sửa lại đề xíu, có lẽ bn chép sai ở phân số cuối của D phải là 1/101
C = 1002+12/100.1 + 992+22/99.2 + ... + 512+502/51.50
C = 1002/100.1 + 12/100.1 + 992/99.2 + 22/99.2 + ... + 512/51.50 + 502/51.50
C = 100/1 + 1/100 + 99/2 + 2/99 + ... + 51/50 + 50/51
C = 100/1 + 99/2 + 98/3 + ... + 51/50 + 50/51 + ... + 1/100
C = (1 + 1 + ... + 1) + 99/2 + 98/3 + ... + 1/100
100 số 1
C = (99/2 + 1) + (98/3 + 1) + ... + (1/100 + 1) + 1
C = 101/2 + 101/3 + ... + 101/100 + 101/101
C = 101.(1/2 + 1/3 + ... + 1/100 + 1/101)
=> C : D = 101
E = \(\frac{100^2+1^2}{100.1}\) + \(\frac{99^2+2^2}{99.2}\) + ...... + \(\frac{51^2+50^2}{51.50}\)
F = \(\frac{1}{2}\)+ \(\frac{1}{3}\)+ ..... + \(\frac{1}{101}\)
G = \(\frac{1}{100.1}\)+ \(\frac{1}{99.2}\)+ .... + \(\frac{1}{51.50}\)
Tính tỉ số của E với F
Tính F - 101 . G
