A=1/11+1/12+1/13+...+1/30
Tìm A
\( A=\frac{1+13+13^2+...+13^{13}}{1+13+13^2+...+13^{12}}\) và \(B=\frac{1+11+11^2+...+11^{13}}{1+11+11^2+...+11^{12}}\) chứng minh A > B
CTR:
a, 1/20+1/21+1/22+...+1/49<13/12
b, 1/11+1/12+1/13+...+1/40>13/12
A = 7/12 + 5/12 : 6 - 11/36
B= ( 4/5 + 1/2 ) : ( 3/13 - 8/13 )
C = ( 2/3 - 1/4 + 5/11) : ( 5/12 + 1 - 7/11 )
A) 7/12+5/12 : 6 -11/36
=7/12 + 5/72 -11/36
=47/72-11/36
=25/72
Tính nhanh
a) 1/10*11 + 1/11*12 + 1/12*13 + 1/13*14 +.........+ 1/78*79
b) 8/7*9 + 8/9*11 + 8/11*13 + 8/13*14 +............+ 8/133*135
c) 12/8*11 + 12/11*14 + 12/14*17 +.........+ 12/503*506
d) 1/4*7 + 1/7*10 + 1/10*13 + 1/13*16 +........+ 1/391*394
e) 4/5*8 + 4/8*11 + 4/11*14 + 4/14*17 +.........+ 4/602*605
g) 1 + 1/3 + 1/6 + 1/10 + 1/15 +...........+ 1/802
Các bạn làm bài tập này giúp mình với ạ
bạn nào làm xong trước và đúng thì mình tick cho nhé
Dấu này * là dấu nhân
Một năm rồi không có ai trả lời à
THấy cx thương nhưng mk nhìn cái đề thì dài thật cx khó có ai có thời gian mà giải
Tính nhanh:
A = 1/10 + -1/11 + 1/12 + -1/13 + 1/14 + -1/15 + 1/16 +1/15 + -1/14 + 1/13 + -1/12 + 1/11 + -1/10
(1/10+-1/10)+(1/11+-1/11)+(1/12+-1/12)+(-1/13+1/13)+(-1/14+1/14)+(-1/15+1/15)+1/16
=0 + 0 +0 + 0 +0 +0 +1/16
=1/16
cho A= 1/11+1/12+1/13+...+1/100 CM: 9/7<A<13/4
TÍNH NHANH
a) 1/10*11+1/11*12+1/12*13+1/13*14+........+1/78*79
b) 8/7*9+8/9*11+8/11*13+8/13*15+.......+8/133*135
c) 12/8*11+12/11*14+12/14*17+.........+12/503*506
d) 1/4*7+1/7*10+1/10*13+1/13*16+..........+1/391*394
e) 4/5*8+4/8*11+4/11*14+4/14*17+.........+4/602*605
g) 1+1/3+1/6+1/10+1/15+........+1/802
DẤU NÀY * LÀ DẤU NHÂN
CÁC BẠN GIẢI GIÚP MÌNH VỚI Ạ
MÌNH CHỈ CÒN KHOẢNG 15 PHÚT NỮA THÔI CẦU XIN CÁC BẠN ĐÓ
a) = 1/10 - 1/11 + 1/11 -1/12 + 1/12 - 1/13 +1/13 1/14 +...+ 1/78 - 1/79
= 1/10 - 1/79
= máy tính ok
mấy câu khác bn làm tương tự là đc nhưng nhớ nhanh thêm khoảng cách giữa các mẫu nha
a)\(\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{78.79}=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{78}-\frac{1}{79}=\frac{1}{10}-\frac{1}{79}=\frac{69}{790}\)
b) \(\frac{8}{7.9}+\frac{8}{9.11}+...+\frac{8}{133.135}=4\left(\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{133.135}\right)\)
\(=4\left(\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{133}-\frac{1}{135}\right)=4\left(\frac{1}{7}-\frac{1}{135}\right)=4.\frac{128}{945}=\frac{456}{945}\)
c) \(\frac{12}{8.11}+\frac{12}{11.14}+...+\frac{12}{503.506}=4\left(\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{503.506}\right)\)
\(=4\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{503}-\frac{1}{506}\right)=4\left(\frac{1}{8}-\frac{1}{506}\right)=\frac{249}{506}\)
d) \(\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{391.394}=\frac{1}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{391.394}\right)\)
\(=\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{391}-\frac{1}{394}\right)=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{394}\right)=\frac{1}{3}.\frac{195}{788}=\frac{65}{788}\)
e) \(\frac{4}{5.8}+\frac{4}{8.11}+...+\frac{4}{602.605}=\frac{4}{3}.\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{602.605}\right)\)
\(=\frac{4}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{602}-\frac{1}{605}\right)=\frac{4}{3}\left(\frac{1}{5}-\frac{1}{605}\right)=\frac{4}{3}.\frac{24}{121}=\frac{32}{121}\)
g) Sửa đề\(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{820}=2\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{1640}\right)=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{40.41}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{40}-\frac{1}{41}\right)=2\left(1-\frac{1}{41}\right)=2.\frac{40}{41}=\frac{80}{41}\)
Bài làm:
a) \(\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{78.79}\)
\(=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{78}-\frac{1}{79}\)
\(=\frac{1}{10}-\frac{1}{79}=\frac{69}{790}\)
b) \(\frac{8}{7.9}+\frac{8}{9.11}+...+\frac{8}{133.135}\)
\(=4\left(\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{133.135}\right)\)
\(=4\left(\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{133}-\frac{1}{135}\right)\)
\(=4\left(\frac{1}{7}-\frac{1}{135}\right)\)
\(=4.\frac{128}{945}=\frac{512}{945}\)
c) \(\frac{12}{8.11}+\frac{12}{11.14}+...+\frac{12}{503.506}\)
\(=4\left(\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{503.506}\right)\)
\(=4\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{503}-\frac{1}{506}\right)\)
\(=4\left(\frac{1}{8}-\frac{1}{506}\right)\)
\(=4.\frac{249}{2024}=\frac{249}{506}\)
d) \(\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{391.394}\)
\(=\frac{1}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{391.394}\right)\)
\(=\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{391}-\frac{1}{394}\right)\)
\(=\frac{1}{3}\left(\frac{1}{4}-\frac{1}{394}\right)\)
\(=\frac{1}{3}.\frac{195}{788}=\frac{65}{788}\)
e) \(\frac{4}{5.8}+\frac{4}{8.11}+...+\frac{4}{602.605}\)
\(=\frac{4}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{602.605}\right)\)
\(=\frac{4}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{602}-\frac{1}{605}\right)\)
\(=\frac{4}{3}\left(\frac{1}{5}-\frac{1}{605}\right)\)
\(=\frac{4}{3}.\frac{24}{121}=\frac{32}{121}\)
g) Phải sửa \(\frac{1}{802}\) thành \(\frac{1}{820}\) nhé
\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{820}\)
\(=1+\frac{1}{1.3}+\frac{1}{2.3}+\frac{1}{2.5}+...+\frac{1}{41.20}\)
\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{40.41}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{40}-\frac{1}{41}\right)\)
\(=2\left(1-\frac{1}{41}\right)\)
\(=2.\frac{40}{41}=\frac{80}{41}\)
Tính : a 1/10×11 + 1/ 11×12 +1/12×13 + .... +1/99×100
b 1/ 1×3 + 1/ 3 ×5 +1/5×7 + .... + 1/97×99
a, 13/19 + 1 - 15/19 - 4/19
b, 3/5 +6/11 +7/13 +2/5 +16/11 +19/13
c, 1/3 +1/6 + 1/12 +1/24 +1/48
d, 1/2 +1/6 +1/12 +1/20 +1/30 +1/42
cho A=1/11+1/12+1/13+...+1/70 CM rằng A<3,5