Thực hiện phép tính
10x11+11x12+12x13+.....+29x30
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\(\frac{5}{10.11}+\frac{5}{11.12}+\frac{5}{12.13}+....+\frac{5}{49.50}\)
\(=5.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+.....+\frac{1}{49.50}\right)\)
\(=5.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{49}-\frac{1}{50}\right)\)
\(=5.\left(\frac{1}{10}-\frac{1}{50}\right)\)
\(=5.\frac{2}{25}\)
\(=\frac{2}{5}\)
5/10 x 11 + 5/11 x 12 + 5/12 x 13 + .... + 5/49 x 50
= 5/10 - 5/11 + 5/11 - 5/12 + 5/12 - 5/13 + ....... + 5/49 - 5/50
= 5/10 - 5/50
= 2/5
S = 1x2 + 2x3 + 3x4 + ……………… + 11x12 + 12x13
3S=1x2x3 + 2x3x3 + 3x4x3+ ………. + 11x12x3 + 12x13x3
Ta lấy K = 1x2x3 +2x3x4 + 3x4x5 + …… + 11x12x13 + 12x13x14
- 3S = 1x2x3 + 2x3x3 + 3x4x3+ ……… + 11x12x3 + 12x13x3
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K – 3S = 0 + 2x3x1 + 3x4x2 + …… .. + 11x12x10 + 12x13x11
K – 3S = K – 12x13x14
Từ đó suy ra: 3S = 12x13x14
S = 4x13x14 = 728
Cách 2:
S x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + …. + 11x12x(13-10) + 12x13x(14-11)
S x 3 = 1x2x3 + 2x3x4 – 2x3x1 + 3x4x5 – 3x4x2 + …..+ 11x12x13 – 11x12x10 +12x13x14 – 12x13x11
S x 3 = 12 x 13 x14
S = 4 x 13 x 14
S = 728
ai k minh minh k lai cho
C=7/10x11+7/11x12+7/12x13+.................+7/69x70
C=1x7/10x11+1x7/11x12+...........+1x7/69x70
C=7(1/10x11+1/11x12+1/12x13+....+1/69x70)
C=7(1/10-1/11+1/11-1/12+1/12-1/13+.......+1/69-1/70)
C=7(1/10-1/70)
C=7(7/70-1/70)
C=7x6/70
C=3/5
`A=7/(10*11)+7/(11*12)+7/(12*13)+...+7/(69*70)`
`1/7A=1/(10*11)+1/(11*12)+1/(12*13)+..+1/(69*70)`
`1/7A=1/10-1/11+1/11-1/12+1/12-1/13+...+1/69-1/70`
`1/7A=1/10-1/70`
`1/7A=7/70-1/70`
`1/7A=6/70`
`A=3/5`
\(A=\dfrac{7}{10.11}+\dfrac{7}{11.12}+\dfrac{7}{12.13}+...+\dfrac{7}{69.70}\)
\(A=7.\left(\dfrac{1}{10.11}+\dfrac{1}{11.12}+\dfrac{1}{12.13}+...+\dfrac{1}{69.70}\right)\)
\(A=7\left(\dfrac{11-10}{10.11}+\dfrac{12-11}{11.12}+\dfrac{13-12}{12.13}+...+\dfrac{70-69}{69.70}\right)\)
\(A=7.\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+...+\dfrac{1}{69}-\dfrac{1}{70}\right)\)
\(A=7.\left(\dfrac{1}{10}-\dfrac{1}{70}\right)\)
\(A=7.\dfrac{3}{35}=\dfrac{3}{5}\)
\(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(=7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)
\(=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(=7.\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(=7.\frac{3}{35}=\frac{3}{5}\)
\(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(A=1\left(\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\right)\)
\(A=7\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)
\(A=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(A=7\left(\frac{1}{10}-\frac{1}{70}\right)=7.\frac{3}{35}=\frac{3}{5}\)
C=7/10x11+7/11x12+7/12x13+.................+7/69x70
C=1x7/10x11+1x7/11x12+...........+1x7/69x70
C=7(1/10x11+1/11x12+1/12x13+....+1/69x70)
C=7(1/10-1/11+1/11-1/12+1/12-1/13+.......+1/69-1/70)
C=7(1/10-1/70)
C=7(7/70-1/70)
C=7x6/70
C=3/5
\(\frac{5}{11x12}+\frac{5}{12x13}+...+\frac{5}{98x99}\)
=\(\frac{5}{11}-\frac{5}{12}+\frac{5}{12}-\frac{5}{13}+...+\frac{5}{98}-\frac{5}{99}\)
=\(\frac{5}{11}-\frac{5}{99}\)
=\(\frac{40}{99}\)
Cái cuối bỏ 1 số 0 thì đúng hơn nha bạn
\(\frac{5}{11.12}+\frac{5}{12.13}+\frac{5}{13.14}+...+\frac{5}{98.99}\)
\(=5\left(\frac{1}{11.12}+\frac{1}{12.13}+\frac{1}{13.14}+...+\frac{1}{98.99}\right)\)
\(=5\left(\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{98}-\frac{1}{99}\right)\)
\(=5\left(\frac{1}{11}-\frac{1}{99}\right)\)
\(=5.\frac{8}{99}\)
\(=\frac{40}{99}\)