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1/4*9+1/9*14+1/14*19+...+1/1999*2004
= 1/5 (1/4 - 1/9 + 1/9 - 1/14 + ... + 1/1999 - 1/2004)
= 1/5 (1/4 - 1/2004)
= 1/5 (501/2004 - 1/2004)
= 1/5 . 125/501
= 25/501
\(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\frac{1}{14\cdot19}+...+\frac{1}{1999+2004}\).
Có sai đề không vậy???
Sửa đề một chút :v
\(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\frac{1}{14\cdot19}+...+\frac{1}{1999\cdot2004}\)
\(=\frac{1}{5}\left[\frac{5}{4\cdot9}+\frac{5}{9\cdot14}+\frac{5}{14\cdot19}+...+\frac{5}{1999\cdot2004}\right]\)
\(=\frac{1}{5}\left[\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{1999}-\frac{1}{2004}\right]\)
\(=\frac{1}{5}\left[\frac{1}{4}-\frac{1}{2004}\right]\)
\(=\frac{1}{5}\cdot\frac{125}{501}=\frac{25}{501}\)
Đặt A =1/4 x 9 + 1/9 x 14 + 1/14 x 19 +...+ 1/1999 + 2004. Ta có:
A= 1/4 x 9 + 1/9 x 14 + 1/14 x 19 +...+ 1/1999 + 2004
5A= 5/4 x 9 + 5/9 x 14 + 5/14 x 19 +...+ 5/1999 + 2004
5A= 1/4 - 1/9 + 1/9 - 1/14 + 1/14 - 1/19 +...+ 1/1999 - 1/2004
5A= 1/4 - 1/2004
A= (1/4 - 1/2004)/5
a)=(3/8+10/16)+(7/12+10/24)
=1+1=2
c)=(4/6+14/6)+(7/13+19/13)+(17/9+1/9)
=3+2+2=7
(19/13+7/13)+(14/6+4/6)+(1/9+17/9)
=26/13+18/6+18/9
=2+3+2
=7
A = \(\dfrac{4}{3}\) + \(\dfrac{17}{3}\) + \(\dfrac{17}{9}\) + \(\dfrac{19}{13}\) + \(\dfrac{1}{9}\) + \(\dfrac{14}{6}\)
A = (\(\dfrac{4}{3}\) + \(\dfrac{17}{3}\)) + (\(\dfrac{17}{9}\) + \(\dfrac{1}{9}\)) + (\(\dfrac{19}{13}\) + \(\dfrac{14}{6}\))
A = \(\dfrac{21}{3}\) + \(\dfrac{18}{9}\) + \(\dfrac{148}{39}\)
A = 7 + 2 + \(\dfrac{148}{39}\)
A = 9 + \(\dfrac{148}{39}\)
A = \(\dfrac{39\times9}{39}\) + \(\dfrac{148}{39}\)
A = \(\dfrac{499}{39}\)
2:
b=2000*2004
=(2002-2)*(2002+2)
=2002^2-4
=>b<a
1:
a: \(=8\cdot9\left(14+17+19\right)=72\cdot50=3600\)
Bài 1:
\(8\times9\times14+6\times17\times12+19\times4\times18\)
\(=8\times9\times14+3\times2\times17\times2\times2\times3+19\times4\times2\times9\)
\(=8\times9\times14+17\times8\times9+19\times8\times9\)
\(=8\times9\times\left(14+17+19\right)\)
\(=8\times9\times50\)
\(=72\times5\times10\)
\(=360\times10\)
\(=3600\)
Bài 2:
Ta có:
\(a=2022\times2022\)
Và: \(b=2000\times2004\)
Mà: \(2022>2000,2022>2004\)
\(\Rightarrow2022\times2022>2000\times2004\)
\(\Rightarrow a>b\)
\(\frac{1}{4\times9}+\frac{1}{9\times14}+\frac{1}{14\times19}+...+\frac{1}{1999\times2004}\)
\(=\frac{1}{5}\times\left(\frac{5}{4\times9}+\frac{5}{9\times14}+\frac{5}{14\times19}+...+\frac{5}{1999\times2004}\right)\)
\(=\frac{1}{5}\times\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{1999}-\frac{1}{2004}\right)\)
\(=\frac{1}{5}\times\left(\frac{1}{4}-\frac{1}{2004}\right)\)
\(=\frac{1}{5}\times\frac{500}{2004}=\frac{25}{501}\)