B=1/30+1/42+1/56+1/72+1/90+1/110+1/132+1/156
b = 1/30 + 1/42+ 1/56+1/72+1/90+1/110+1/132+1/156
B = 1/30 + 1/42 + 1/56 + 1/72 + 1/90 + 1/110 + 1/132 + 1/156
B = 1/5x6 + 1/6x7 + 1/7x8 + 1/8x9 + 1/9x10 + 1/10x11 + 1/11x12 + 1/12x13
B = 1/5 -1/6 + 1/6- 1/7 + 1/7 -1/ 8 + 1/8 -1/9 +1/9 -1/10 + 1/10 - 1/11 + 1/11-1/12 +1/12 -1/13
B = 1/5 - 1/13
B = 8/65
1/20+1/30+1/42+1/56+1/72+1/90+1/110+1/132+1/156 = ?
1/20 + 1/30 + 1/42 + ... + 1/156
= 1/4.5 + 1/5.6 + 1/6.7 + .... + 1/12.13
= 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + ... + 1/12 - 1/13
= 1/4 - 1/13
= 9/52
\(=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{12}-\frac{1}{13}\)
\(=\frac{1}{4}-\frac{1}{13}=\frac{9}{52}\)
****
= 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8 + 1/8.9 + 1/9.10 + 1/10.11 + 1/11.12 + 1/12.13
= 1/4 - 1/5 + 1/5 - 1/6 + ... + 1/2 - 1/13
= 1/4 - 1/13
= 9/52
Tính 1/20 + 1/30 + 1/42 + 1/56 + 1/ 72 + 1/90 + 1/110 + 1/ 132 + 1/156
Đặt tổng trên là A ta có :
A= 1/ 20 + 1/ 30 + 1/ 42 + 1/ 56 + 1/ 72 + 1/90 + 1/110 + 1 / 123 + 1/ 156
= 1 / 4 x5 + 1/ 5 x 6 + 1/6x 7 + 1/ 7x8 + 1/8x9 + 1/9x10+ 1/ 10x11+ 1 /11x12 +1/12 x 13
= 1/4- 1/5 + 1/ 5 - 1/6 + 1/ 6 - 1/7 + 1/ 7 - 1/8 + 1/8 - 1/9 + 1/9 - 1/10+ 1/10 - 1/11 + 1/11 - 1/12+ 1/ 12 - 1/13
= 1 /4 - 1 /13
= 9 /52
A = 1/20 + 1/30 + 1/42 + 1/56 + 1/72 + 1/90 + 1/ 110 + 1/132 + 1/156
\(=\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{12.13}\)
áp dụng \(\frac{1}{a.b}=\frac{1}{a}-\frac{1}{b}\)làm sẽ có các số nghịch đảo và được kết quả là 1/4 - 1/13
A = 1/20 + 1/30 + 1/42 + 1/56 + 1/72 + 1/90 + 1/110 + 1/132 + 1/156
A = 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8 + 1/8.9 + 1/9.10 + 1/10.11 + 1/11.12 + 1/12.13
A = 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + 1/9 - 1/10 + 1/10 - 1/11 + 1/11 - 1/12 + 1/12 - 1/13
A = 1/4 - 1/13
A = 9/52
A = \(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{132}+\frac{1}{156}\)
= \(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{11.12}+\frac{1}{12.13}\)
= \(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}\)
= \(\frac{1}{4}-\frac{1}{13}\)
= \(\frac{9}{52}\)
Vậy \(A=\frac{9}{52}\)
A=1/42+1/56+1/72+1/90+1/110+1/132+1/156+1/182+1/210
A = \(\frac{1}{10}=0,1\)
Tích cho tớ để đủ 250 điểm đi
=> A = 1/6*7 + 1/7*8 + 1/8*9 + 1/9*10 + ... + 1/13*14 + 1/14*15 = 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + 1/9 - 1/10 +...+ 1/13 - 1/14 + 1/14 - 1/15 = 1/6 - 1/15 =1/10 Nhớ ^^
tinh:1/42+1/56+1/72+1/90+1/110+1/132+1/156+1/182+1/210
=1/6*7+1/7*8+1/8*9...+1/14*15
=1/6-1/7+1/7-1/8+...+1/14-1/15
=1/6-1/15
=1/10
1/42 + 1/56 + 1/72 + 1/90 + 1/110 + 1/132 + 1/156 + 1/182 + 1/210
= 1/6.7 + 1/7.8 + 1/8.9 + 1/9.10 + 1/10.11 + 1/11.12 + 1/12.13 + 1/13.14 + 1/14.15
= 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + 1/9 - 1/10 + 1/10 - 1/11 + 1/11 - 1/12 + 1/12 - 1/13 + 1/13 - 1/14 + 1/14 - 1/15
= 1/6 - 1/15
= 1/10
\(=\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{14.15}\)
\(=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{14}-\frac{1}{15}\)
\(=\frac{1}{6}-\frac{1}{15}\)
\(=\frac{1}{10}\)
B= 1/30 + 1/42 + 1/56 + 1/72 + 1/90 + 1/110 + 1/132
\(\frac{1}{5x6}+\frac{1}{6x7}+\frac{1}{7x8}+\frac{1}{8x9}+\frac{1}{9x10}+\frac{1}{10x11}+\frac{1}{11x12}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}\)
=1/5-1/12=7/60
B= 1/30 + 1/42 + 1/56 + 1/72 + 1/90 + 1/110 + 1/132= 7/60
\(A=\frac{1}{5}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}+\frac{1}{156}\)
A = 1/5 + [1/5.6 + 1/6.7 + ... + 1/12.13]
A = 1/5 + [1/5-1/6+1/6-1/7+...+1/12-1/13]
A = 1/5 + [1/5-1/13]
A = 1/5 + 8/65
A = 21/65
1/30+1/42+1/56+1/72+1/90+1/110+1/132
1/30 + 1/42 +1/56 +1/72+1/90+1/110+1/132
= 1/5x6+1/6x7+1/7x8+1/8x9+1/9x10+1/10x11+1/11x12
=1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10+1/10-1/11+1/11-1/12
= 1/5 -1/12
=7/60
làm sai rồi lấy đâu ra dấu trừ phải là 1/5+ 1/12 chứ