Tính \(N=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{3000}}{\frac{2999}{1}+\frac{2998}{2}+\frac{2997}{3}+...+\frac{1}{2999}}\)
A=
\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{3000}\)
\(\frac{2999}{1}+\frac{2998}{2}+\frac{2997}{3}+...+\frac{1}{2999}\)
Tính A = \(\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{3000}}{\frac{2999}{1}+\frac{2998}{2}+...+\frac{1}{2999}}\)(Trình bày rõ => tick )
Ta có \(A=\frac{\frac{1}{2}+\frac{1}{3}+....+\frac{1}{3000}}{\frac{2999}{1}+\frac{2998}{2}+...+\frac{1}{2999}}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{3000}}{\left(1+1+...+1\right)+\frac{2998}{2}+...+\frac{1}{2999}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{3000}}{\left(1+\frac{2998}{2}\right)+\left(1+\frac{2997}{3}\right)+...+\left(1+\frac{1}{2999}\right)+\frac{3000}{3000}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{3000}}{\frac{3000}{2}+\frac{3000}{3}+...+\frac{3000}{3000}}\)
= \(\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{3000}}{3000\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{3000}\right)}=\frac{1}{3000}\)
Vậy A= \(\frac{1}{3000}\)
Ai đó giúp tui đi , sáng mai kiểm tra ròi :'(
Bài 1 : Tính A = \(\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{3000}}{\frac{2999}{1}+\frac{2998}{2}+...+\frac{1}{2999}}\)(Giải rõ => tick )
Câu 1 : Tính
A= 1/2 + 1/3 + 1/4 +...+1/300
B= 2999/1 + 2998/2 + 2997/3 +...+1/2999
Tính \(\frac{A}{B}\)
Câu 2
C= (1+2012/1)(1+2012/2)....(1+2012/1000)
D=(1+1000/1)(1+1000/2)(1+1000/3)...(1+1000/2012)
Tính \(\frac{C}{D}\)
Câu 3
Cho E=1/1.2 + 1/3.4 + 1/5.6 +...+1/2013/2014
F=1/1008/2014 + 1009/2013 +.....+1/2014.1008
Tính \(\frac{E}{F}\)
Câu 1:
B = \(\frac{2999}{1}+\frac{2998}{2}+\frac{2997}{3}+...+\frac{1}{2999}\)
= \(\frac{3000-1}{1}+\frac{3000-2}{2}+\frac{3000-3}{3}+...+\frac{3000-2999}{2999}\)
= \(\left(\frac{3000}{1}+\frac{3000}{2}+\frac{3000}{3}+...+\frac{3000}{2999}\right)-\left(\frac{1}{1}+\frac{2}{2}+\frac{3}{3}+...+\frac{2999}{2999}\right)\)
= \(3000+3000.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2999}\right)-2999\)
= \(3000\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2999}\right)+\frac{3000}{3000}\)
= \(3000\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{3000}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{3000}}{3000\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{3000}\right)}=\frac{1}{3000}\)
C=2013/1*2014/2*2015/3*...*3012/1000
C=2013*2014*2015*...*3012/1*2*3*...*1000
D=1001/1*1002/2*1003/3*...*3012/2012
D=1001*1002*...*3012/1*2*...*2012
Suy ra C/D=2013*2014*2015*...3012*1*2*...*2012/1*2*3*...*1000*1001*1002*...*3012
( Nhân đảo ngược)
Vậy C/D=1
Cho N = 30 /1 + 30 / 2 + 30 / 3 +.....+ 30 / 3000 và M = 2999 / 1 + 2998 / 2 + 2997 / 3 + ...+ 1 / 2999
Tính N : M
1/2+1/3+1/4+...+1/300
___________________
2999/1+2998/2+2997/3+...+1/2999
Đề là 1/3000 nhé ~
\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{3000}}{\frac{2999}{1}+\frac{2998}{2}+\frac{2997}{3}+...+\frac{1}{2999}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{3000}}{\left(\frac{2998}{2}+1\right)+\left(\frac{2997}{3}+1\right)+...+\left(\frac{1}{2999}+1\right)+1}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{3000}}{\frac{3000}{2}+\frac{3000}{3}+....+\frac{3000}{2999}+\frac{3000}{3000}}\)
\(=\frac{1}{3000}\)
1, Tính \(\frac{1}{2}-\left(\frac{1}{3}+\frac{2}{3}\right)+\left(\frac{1}{4}+\frac{2}{4}+\frac{3}{4}\right)-\left(\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}\right)+...+\left(\frac{1}{100}+\frac{2}{100}+\frac{3}{100}+...+\frac{99}{100}\right)\)2,Tính \(\left(1-\frac{1}{2^2}\right)x\left(1-\frac{1}{3^2}\right)x\left(1-\frac{1}{4^2}\right)x...x\left(1-\frac{1}{n^2}\right)\)
tính biểu thức
\(\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}}{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}}:\frac{3+\frac{3}{2}+\frac{3}{3}+\frac{3}{4}}{2-\frac{2}{2}+\frac{2}{3}-\frac{2}{4}}\)
giải:
ta có :
\(\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}}{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}}:\frac{3+\frac{3}{2}+\frac{3}{3}+\frac{3}{4}}{2-\frac{2}{2}+\frac{2}{3}-\frac{2}{4}}\)
\(\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}}{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}}.\frac{2\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}{3\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)}=\frac{2}{3}\)
Tính nhah ---- giúp mik giải nâ các bn thank nhiều nhiều
a)\(\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}}{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}}:\frac{3+\frac{3}{2}+\frac{3}{3}+\frac{3}{4}}{2-\frac{2}{2}+\frac{2}{3}-\frac{2}{4}}+\frac{1}{3}\)
b) \(\frac{\frac{1}{3}-\frac{1}{5}-\frac{1}{7}}{\frac{2}{3}-0,4-\frac{2}{7}}+\frac{\frac{3}{8}-\frac{3}{16}-\frac{3}{32}+\frac{3}{64}}{\frac{1}{4}-\frac{1}{8}-\frac{1}{16}+\frac{1}{32}}\)
c) \(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}\)