(1/(1x2)/(2x3)/(3x4)):(1/(2x3)/(3x4)/(4x5)):...(1/(97*98)/(98*99)/(99*100
Những câu hỏi liên quan
(1/(1x2)/(2x3)/(3x4)):(1/(2x3)/(3x4)/(4x5)):...(1/(97*98)/(98*99)/(99*100))
1/(1x2)/(2x3)/(3x4)):(1/(2x3)/(3x4)/(4x5)):...(1/(97*98)/(98*99)/(99*100
(1/(1x2)/(2x3)/(3x4)):(1/(2x3)/(3x4)/(4x5)):...(1/(97*98)/(98*99)/(99*100
Lên Qanda mà hỏi
(1/(1x2)/(2x3)/(3x4)):(1/(2x3)/(3x4)/(4x5)):...(1/(97*98)/(98*99)/(99*100))
1/(1x2)/(2x3)/(3x4)):(1/(2x3)/(3x4)/(4x5)):...(1/(97*98)/(98*99)/(99*100
(1/(1x2)/(2x3)/(3x4)):(1/(2x3)/(3x4)/(4x5)):...(1/(97*98)/(98*99)/(99*100
ai lam dc mik tick cho
tính tổng: 1/1x2 + 1/2x3 + 1/ 3x4 +...+ 1/98 x 99 + 1/99 x 100
1/1*2 + 1/2*3 + 1/3*4 + .... + 1/99 * 100
= 1- 1/100
= 99/100
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=1-1/2+1/2-...-1/99+1/99-1/100=1-1/100=99/100
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Calculate: 1/2x3+1/3x4 + 1/4x5+...+1/98 x 99 = A /198 Answer: A =
\(\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{98x99}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}\)
\(=\frac{1}{2}-\frac{1}{99}\)
\(=\frac{99}{198}-\frac{2}{198}\)
\(=\frac{97}{198}\)
\(\frac{A}{198}=\frac{97}{198}=>A=198x97:198=97\)
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S=1x2 + 2x3 + 3x4 + 4x5 + ... + 99