Tính nhanh
\(\frac{3}{1\cdot3}+\frac{3}{3\cdot5}+\frac{3}{5\cdot7}+...+\frac{3}{99\cdot100}\)
Giúp mk với T.T
tính rồi so sánh m và n biết
m = \(\frac{3}{1\cdot3}\)+ \(\frac{3}{3\cdot5}\)+\(\frac{3}{5\cdot7}\)+...+\(\frac{2}{99\cdot100}\)
n = \(\frac{3}{1\cdot3}\)+ \(\frac{3}{3\cdot5}\)+ \(\frac{3}{5\cdot7}\)+ ...+ \(\frac{3}{97\cdot99}\)
rồi, mấy bạn giải ra giùm mk nhé. 4h mk di học rồi cho nên trước 4 giờ các bạn giải cho mk nhé.
n=\(\frac{2}{3}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
n=\(\frac{2}{3}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
n=\(\frac{2}{3}\left(1-\frac{1}{99}\right)\)
n=\(\frac{2}{3}\times\frac{98}{99}\)
n=\(\frac{196}{297}\)
Câu \(M=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{2}{99.100}\)Bạn viết \(\frac{3}{99.100}=\frac{2}{99.100}\)mik sửa lại nhé.
\(M=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{99.100}\)
\(M=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{100-99}{99.100}\)
\(M=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(M=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{100}\right)\)
\(M=\frac{3}{2}.\frac{99}{100}=\frac{297}{200}\)
\(N=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+....+\frac{3}{97.99}\)
\(N=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+....+\frac{99-97}{97.99}\)
\(N=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{97}-\frac{1}{99}\right)\)
\(N=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{99}\right)\)
\(\Rightarrow N=\frac{3}{2}.\frac{98}{99}=\frac{49}{33}\)
Ta thấy : \(\frac{297}{200}>\frac{49}{33}\Rightarrow M>N\)
tính nhanh
a, \(\frac{2}{2\cdot3}\)+\(\frac{2}{3\cdot4}\)+\(\frac{2}{4\cdot5}\)+..........+\(\frac{2}{99\cdot100}\)
b.\(\frac{1}{1\cdot3}\)+ \(\frac{1}{3\cdot5}\)+ \(\frac{1}{5\cdot7}\)+\(\frac{1}{99\cdot101}\)
ai nhanh nhất mink tick và số lượng tick có hạn chỉ trong tối nay vì mai mink phải nộp bài
tính hợp lí :
B=\(\frac{1\cdot4}{2\cdot3}+\frac{2\cdot5}{3\cdot4}+\frac{3\cdot6}{4\cdot5}+.....+\frac{98\cdot101}{99\cdot100}\)
\(\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+\frac{4}{7\cdot9\cdot11}+\frac{4}{9\cdot11\cdot13}\)
giúp mk nha các bn
\(\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+\frac{4}{7\cdot9\cdot11}+\frac{4}{9\cdot11\cdot13}\)
\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)
\(=\frac{1}{1.3}-\frac{1}{11.13}\)
\(=\frac{1}{3}-\frac{1}{143}\)
\(=\frac{140}{429}\)
\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+.....+\frac{1}{99\cdot100}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+........+\frac{1}{99.100}\\ =\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+.........+\left(\frac{1}{99}-\frac{1}{100}\right)\\ =\frac{1}{2}+\left(\frac{1}{3}-\frac{1}{3}\right)+......+\left(\frac{1}{99}-\frac{1}{99}\right)-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\\ =\frac{49}{100}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+...+\frac{1}{99}.\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+..+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{49}{100}\)
Hãy tính tổng: \(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\)
Ta có: 2/1.3 = 1/1 - 1/3
2/3.5 = 1/3 - 1/5
\(\Rightarrow\) 2/1.3 + 2/3.5 + 2/5.7 + ... + 2/99.101
= 1/1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/99 - 1/100
= 1 - 1/100
= 99/100
tích trên sẽ = 1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/100
=1-1/100 =99/100
bạn nhớ rằng k/n.(n+k) sẽ = 1/n-1/n+k
=1-1/3+1/3-1/5+1/5-1/7+....+1/99-1/101
=1-1/101
=100/101
Đúng 100 phần trăm luôn
\(\frac{\frac{3}{5}\cdot7^2-3,5^6+\frac{3}{5}\cdot3^9}{\frac{3}{4}\cdot7^2-\frac{3}{4}\cdot5^7+\frac{3}{4}\cdot3^9}\)
Cho \(U_1=\frac{1}{1\cdot3\cdot5}\); \(U_2=\frac{1}{1\cdot3\cdot5}+\frac{1}{3\cdot5\cdot7}\); \(U_3=\frac{1}{1\cdot3\cdot5}+\frac{1}{3\cdot5\cdot7}+\frac{1}{5\cdot7\cdot9}\)
a) Lập quy trình tổng quát tính Un
b) Tính U50, U60
c) Tính U1002
bài 1 tính tổng
a) \(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\)
b) \(\frac{5}{1\cdot3}+\frac{5}{3\cdot5}+\frac{5}{5\cdot7}+...+\frac{5}{99\cdot101}\)
bài 2 chứng tỏ rằng phân số \(\frac{2n+1}{3n+2}\)là phân số tối giản.
bài 3 cho A=\(\frac{n+2}{n-5}\)(n thuộc z;n khác 5) tìm x để A thuộc z
bài 4 tính giá trị biểu thức
A=\(10101\cdot\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{3\cdot7\cdot11\cdot13\cdot37}\right)\)
Bài 1 :
a) =) \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)= \(1-\frac{1}{101}=\frac{100}{101}\)
b) =) \(\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)
=) \(\frac{5}{2}.\frac{100}{101}=\frac{250}{101}\)( theo phần a)
Bài 2 :
-Gọi d là UCLN \(\left(2n+1;3n+2\right)\)( d \(\in N\)* )
(=) \(2n+1⋮d\left(=\right)3.\left(2n+1\right)⋮d\)
(=) \(6n+3⋮d\)
và \(3n+2⋮d\left(=\right)2.\left(3n+2\right)⋮d\)
(=) \(6n+4⋮d\)
(=) \(\left(6n+4\right)-\left(6n+3\right)⋮d\)
(=) \(6n+4-6n-3⋮d\)
(=) \(1⋮d\left(=\right)d\in UC\left(1\right)\)(=) d = { 1;-1}
Vì d là UCLN\(\left(2n+1;3n+2\right)\)(=) \(d=1\)(=) \(\frac{2n+1}{3n+2}\)là phân số tối giản ( đpcm )
Bài 3 :
-Để A \(\in Z\)(=) \(n+2⋮n-5\)
Vì \(n-5⋮n-5\)
(=) \(\left(n+2\right)-\left(n-5\right)⋮n-5\)
(=) \(n+2-n+5⋮n-5\)
(=) \(7⋮n-5\)(=) \(n-5\in UC\left(7\right)\)= { 1;-1;7;-7}
(=) n = { 6;4;12;-2}
Vậy n = {6;4;12;-2} thì A \(\in Z\)
Bài 4:
A = \(10101.\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{3.7.11.13.37}\right)\)
= \(10101.\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{111111}\right)\)
= \(10101.\left(\frac{1}{111111}+\frac{5}{222222}\right)\)= \(10101.\left(\frac{2}{222222}+\frac{5}{222222}\right)\)
= \(10101.\frac{7}{222222}\)( không cần rút gọn \(\frac{7}{222222}\))
= \(\frac{7}{22}\)