\(E=\frac{100^2+1^2}{100.1}+\frac{99^2+2^2}{99.2}+...+\frac{51^2+50^2}{51.50}\)
Tính E
NS
Những câu hỏi liên quan
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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\(A=\frac{100^2+1^2}{100.1}+\frac{99^2+2^2}{99.2}+...+\frac{52^2+49^2}{52.49}+\frac{51^2+50^2}{51.50}\)
\(B=\frac{1}{100.1}+\frac{1}{99.2}+...+\frac{1}{51.50}\)
\(C=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}+\frac{1}{101}\)
a) Tính\(\frac{A}{C}\)
b)Tính C-101B
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
\(E=\frac{100^2+1^2}{100.1}+\frac{99^2+2^2}{99.}+.....+\frac{51^2+50^2}{51.50}\)
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 FTính F - 101 . G
Đọc tiếp
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
ChoEfrac{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}ChoFfrac{1}{2}+frac{1}{3}+...+frac{1}{100}+frac{1}{101}ChoGfrac{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
Đọc tiếp
\(ChoE=\frac{100^2+1^2}{100.1}+\frac{99^2+2^2}{99.2}+\frac{98^2+3^2}{98.3}+...+\frac{52^2+49^2}{52.49}+\frac{51^2+50^2}{51.50}\)
\(ChoF=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}+\frac{1}{101}\)
\(ChoG=\frac{1}{100.1}+\frac{1}{99.2}+\frac{1}{98.3}+...+\frac{1}{52.49}+\frac{1}{51.50}\)
\(a.Tính:\frac{E}{F}\) \(b.Tính:F-101G\)
ChoEfrac{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}ChoFfrac{1}{2}+frac{1}{3}+...+frac{1}{100}+frac{1}{101}ChoGfrac{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
Đọc tiếp
\(ChoE=\frac{100^2+1^2}{100.1}+\frac{99^2+2^2}{99.2}+\frac{98^2+3^2}{98.3}+...+\frac{52^2+49^2}{52.49}+\frac{51^2+50^2}{51.50}\)
\(ChoF=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}+\frac{1}{101}\)
\(ChoG=\frac{1}{100.1}+\frac{1}{99.2}+\frac{1}{98.3}+...+\frac{1}{52.49}+\frac{1}{51.50}\)
\(a.Tính:\frac{E}{F}\) \(b.Tính:F-101G\)
ChoEfrac{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}ChoFfrac{1}{2}+frac{1}{3}+...+frac{1}{100}+frac{1}{101}ChoGfrac{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
Đọc tiếp
\(ChoE=\frac{100^2+1^2}{100.1}+\frac{99^2+2^2}{99.2}+\frac{98^2+3^2}{98.3}+...+\frac{52^2+49^2}{52.49}+\frac{51^2+50^2}{51.50}\)
\(ChoF=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}+\frac{1}{101}\)
\(ChoG=\frac{1}{100.1}+\frac{1}{99.2}+\frac{1}{98.3}+...+\frac{1}{52.49}+\frac{1}{51.50}\)
\(a.Tính:\frac{E}{F}\) \(b.Tính:F-101G\)
\(ChoE=\frac{100^2+1^2}{100.1}+\frac{99^2+2^2}{99.2}+\frac{98^2+3^2}{98.3}+...+\frac{52^2+49^2}{52.49}+\frac{51^2+50^2}{51.50}\)
\(ChoF=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}+\frac{1}{101}\)
\(ChoG=\frac{1}{100.1}+\frac{1}{99.2}+\frac{1}{98.3}+...+\frac{1}{52.49}+\frac{1}{51.50}\)
\(a.Tính:\frac{E}{F}\) \(b.Tính:F-101G\)
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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
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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
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