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
Những câu hỏi liên quan
\(A=\frac{100^2+1^2}{100.1}+\frac{99^2+2^2}{99.2}+...+\frac{52^2+49^2}{52.49}+\frac{51^2+50^2}{51.50}\)
\(B=\frac{1}{100.1}+\frac{1}{99.2}+...+\frac{1}{51.50}\)
\(C=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}+\frac{1}{101}\)
a) Tính\(\frac{A}{C}\)
b)Tính C-101B
Cho \(E=\frac{100^2+1^1}{100.1}+\frac{99^2+2^2}{99.2}+\frac{98^2+3^2}{98.3}+...+\frac{52^2+49^2}{52.49}+\frac{51^2+50^2}{51.50}\)
\(F=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...\frac{1}{101}\)
\(G=\frac{1}{100.1}+\frac{1}{99.2}+...\frac{1}{51.50}\)
a) Tính \(\frac{E}{F}\)
b) Tính F - 101G
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\(E=\frac{100^2+1^2}{100.1}+\frac{99^2+2^2}{99.2}+...+\frac{51^2+50^2}{51.50}\)
Tính E
\(E=\frac{100}{1}+\frac{1}{100}+\frac{99}{2}+\frac{2}{99}+...+\frac{51}{50}+\frac{50}{51}\)
\(E=\left(\frac{99}{2}+1\right)+\left(\frac{98}{3}+1\right)+...+\left(\frac{50}{51}+1\right)+1\)
\(E=\frac{101}{2}+\frac{101}{3}+...+\frac{101}{101}\)
\(E=101.\left(\frac{1}{2}+\frac{1}{3}+..+\frac{1}{101}\right)\)
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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
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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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
rút gọn phân sốfrac{1.3.5...97.99}{51.52.53...99.100}tính A:BAfrac{100^2+1^2}{100.1}+frac{99^2+2^2}{99.2}+frac{98^2+3^2}{98.3}+...+frac{51^2+50^2}{51.50}Bfrac{1}{2}+frac{1}{3}+frac{1}{4}+...+frac{1}{100}
Đọc tiếp
rút gọn phân số
\(\frac{1.3.5...97.99}{51.52.53...99.100}\)
tính A:B
A=\(\frac{100^2+1^2}{100.1}\)+\(\frac{99^2+2^2}{99.2}\)+\(\frac{98^2+3^2}{98.3}\)+...+\(\frac{51^2+50^2}{51.50}\)
B=\(\frac{1}{2}\)+\(\frac{1}{3}\)+\(\frac{1}{4}\)+...+\(\frac{1}{100}\)