Tính nhanh:
a) 3/5+3/5.9+3/9.13+...+3/97.101
b) (1+1/2).(1+1/3).(1+1/4)...(1+1/99)
GIÚP MÌNH VỚI Ạ.CẢM ƠN MỌI NGƯỜI!
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
BF
19 tháng 2 2023
`3/4 + 5/6 = 9/12 + 10/12 = 19/12`
`1/2 + 7/12 = 6/12 + 7/12 = 13/12`
`2/3 xx 3/4 = 2/4 = 1/2`
`7/4 : 2 = 7/4 xx 1/2 = 7/8`
VQ
19 tháng 2 2023
\(a,\dfrac{3}{4}+\dfrac{5}{6}=\dfrac{18}{24}+\dfrac{20}{24}=\dfrac{38}{24}=\dfrac{19}{12}\)
\(b,\dfrac{1}{2}+\dfrac{7}{12}=\dfrac{6}{12}+\dfrac{7}{12}=\dfrac{13}{12}\)
\(c,\dfrac{2}{3}x\dfrac{3}{4}=\dfrac{2}{4}\)
\(d,\dfrac{7}{4}:2=\dfrac{7}{4}x\dfrac{1}{2}=\dfrac{7}{8}\)
6 tháng 9 2018
Làm trc cho 2 câu cuối
c) \(a^2-b^2-4a+4b\)
\(=\left(a+b\right)\left(a-b\right)-4\left(a-b\right)\)
\(=\left(a-b\right)\left[\left(a+b\right)-4\right]\)
d) \(a^2+2ab+b^2-2a-2b+1\)
\(=\left(a+b\right)^2-2\left(a+b\right)+1\)
\(=\left(a+b\right)\left[\left(a+b\right)-2\right]+1\)
NQ
3
7 tháng 8 2016
1/1.5 + 1/5.9 + 1/9.13 + ... + 1/97.101
= 1/4.(4/1.5 + 4/5.9 + 4/9.13 + ... + 4/97.101)
= 1/4.(1 - 1/5 + 1/5 - 1/9 + 1/9 - 1/13 + ... + 1/97 - 1/101)
= 1/4.(1 - 1/101)
= 1/4.100/101
= 25/101
DD
7 tháng 8 2016
\(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+........+\frac{1}{97.101}\)
\(=\frac{1}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+........+\frac{1}{97}-\frac{1}{101}\right)\)
\(=\frac{1}{4}\left(1-\frac{1}{101}\right)\)
\(=\frac{1}{4}.\frac{100}{101}=\frac{25}{101}\)
6
2
H9
HT.Phong (9A5)
CTVHS
16 tháng 6 2023
\(\left(1-\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\cdot...\cdot\dfrac{97}{98}\cdot\dfrac{98}{99}\)
\(=\dfrac{1\cdot2\cdot3\cdot...\cdot98}{2\cdot3\cdot4\cdot...\cdot99}\)
\(=\dfrac{1}{99}\)
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
16 tháng 6 2023
\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}.....\dfrac{99}{98}.\dfrac{100}{99}=\dfrac{100}{2}=50\)
HT
11 tháng 7 2016
\(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{97.101}\)
\(=\frac{1}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{101}\right)\)
\(=\frac{1}{4}.\left(1-\frac{1}{101}\right)\)
\(=\frac{1}{4}.\frac{100}{101}\)
\(=\frac{25}{101}\)
\(a,\dfrac{3}{5}+\dfrac{3}{5\cdot9}+\dfrac{3}{9\cdot13}+....+\dfrac{3}{97\cdot101}\)
\(=\dfrac{3}{4}\cdot\left(\dfrac{4}{5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+....+\dfrac{4}{97\cdot101}\right)\)
\(=\dfrac{3}{4}\cdot\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+....+\dfrac{1}{97}-\dfrac{1}{101}\right)\)
\(=\dfrac{3}{4}\cdot\left(1-\dfrac{1}{101}\right)\)
\(=\dfrac{3}{4}\cdot\dfrac{100}{101}\)
\(=\dfrac{75}{101}\)
\(b,\left(1+\dfrac{1}{2}\right)\cdot\left(1+\dfrac{1}{3}\right)\cdot\left(1+\dfrac{1}{4}\right)\cdot....\cdot\left(1+\dfrac{1}{99}\right)\)
\(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot....\cdot\dfrac{100}{99}\)
\(=\dfrac{100}{2}=50\)
Tính nhanh:
a) \(\dfrac{3}{5}+\dfrac{3}{5.9}+\dfrac{3}{9.13}+...+\dfrac{3}{97.101}\)
= \(\dfrac{3}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{97}-\dfrac{1}{101}\right)\)
= \(\dfrac{3}{4}\left(1-\dfrac{1}{101}\right)\)
= \(\dfrac{3}{4}\times\dfrac{100}{101}\)
= \(\dfrac{75}{101}\)
b) \(\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{4}\right)...\left(\dfrac{1}{98}+1\right)\left(\dfrac{1}{99}+1\right)\)
\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}...\dfrac{99}{98}.\dfrac{100}{99}\)
\(=\dfrac{3.4.5...99.100}{2.3.4...98.99}\)
\(=\dfrac{100}{2}\)
\(=50\)