tính : \(A=\frac{2}{60\cdot63}+\frac{2}{63\cdot66}+...+\frac{2}{117\cdot120}+\frac{2}{2003}\)
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17 tháng 2 2017
Ta có: \(A=\frac{2}{60.63}+\frac{2}{63.66}+...+\frac{2}{117.120}+\frac{2}{2003}\)
\(\Rightarrow A=\frac{2}{3}\left(\frac{3}{60.63}+\frac{3}{63.66}+...+\frac{3}{117.120}\right)+\frac{2}{2003}\)
\(\Rightarrow A=\frac{2}{3}\left(\frac{1}{60}-\frac{1}{63}+\frac{1}{63}-\frac{1}{66}+...+\frac{1}{117}-\frac{1}{120}\right)+\frac{2}{2003}\)
\(\Rightarrow A=\frac{2}{3}\left(\frac{1}{60}-\frac{1}{120}\right)+\frac{2}{2003}\)
\(\Rightarrow A=\frac{2}{3}.\frac{1}{120}+\frac{2}{2003}\)
\(\Rightarrow A=\frac{1}{180}+\frac{2}{2003}\)
\(B=\frac{5}{40.44}+\frac{5}{44.48}+...+\frac{5}{76.80}+\frac{5}{2003}\)
\(\Rightarrow B=\frac{5}{4}\left(\frac{4}{40.44}+\frac{4}{44.48}+...+\frac{4}{76.80}\right)+\frac{5}{2003}\)
\(\Rightarrow B=\frac{5}{4}\left(\frac{1}{40}-\frac{1}{44}+\frac{1}{44}-\frac{1}{48}+...+\frac{1}{76}-\frac{1}{80}\right)+\frac{5}{2003}\)
\(\Rightarrow B=\frac{5}{4}\left(\frac{1}{40}-\frac{1}{80}\right)+\frac{5}{2003}\)
\(\Rightarrow B=\frac{5}{4}.\frac{1}{80}+\frac{5}{2003}\)
\(\Rightarrow B=\frac{1}{64}+\frac{5}{2003}\)
Vì \(\left\{\begin{matrix}\frac{1}{64}>\frac{1}{180}\\\frac{5}{2003}>\frac{2}{2003}\end{matrix}\right.\Rightarrow\frac{1}{64}+\frac{5}{2003}>\frac{1}{180}+\frac{2}{2003}\Rightarrow B>A\)
Vậy A < B
HS
12 tháng 10 2019
\(1-\frac{2}{3.5}-\frac{2}{5.7}-\frac{2}{7.9}-.......-\frac{2}{61.63}-\frac{2}{63.65}\)
=\(-1.\left(\frac{2}{3.5}+\frac{2}{5.7}+......\frac{2}{63.65}\right)+1\)
=\(-1.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.......+\frac{1}{63}-\frac{1}{65}\right)+1\)
=\(-1.\left(\frac{1}{3}-\frac{1}{65}\right)+1\)
=\(-1.\frac{62}{195}+1\)
=\(\frac{-62}{195}+\frac{195}{195}\)
=\(\frac{133}{195}\)
Hok tốt nhé bn
23 tháng 12 2016
Bài 1:
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
\(\Rightarrow P=\frac{1\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}{5\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}-\frac{2\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2002}\right)}{3\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}\)
\(\Rightarrow P=\frac{1}{5}-\frac{2}{3}\)
\(\Rightarrow P=\frac{-7}{15}\)
Vậy \(P=\frac{-7}{15}\)
Bài 2:
Ta có: \(S=23+43+63+...+203\)
\(\Rightarrow S=13+10+20+23+...+103+100\)
\(\Rightarrow S=\left(13+23+...+103\right)+\left(10+20+...+100\right)\)
\(\Rightarrow S=3025+450\)
\(\Rightarrow S=3475\)
Vậy S = 3475
T
23 tháng 12 2016
1. \(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
=> P =\(\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{5\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}-\frac{2\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}{3\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}\)
=> P = \(\frac{1}{5}-\frac{2}{3}\)
P = \(\frac{3}{15}-\frac{10}{15}\)
=> P =\(\frac{-7}{15}\)
2. ta có:
S = 23 + 43 + 63 +...+ 203
=> S = 13 + 10 + 23 + 20 +...+ 103 + 100
=> S = ( 13 + 23+...+ 103 ) + ( 10 + 20 +...+ 100 )
=> S = 3025 + 550
=> S = 3575
Vậy S = 3575
DT
1
13 tháng 4 2019
Bài 2:
Ta có: A=\(2\left(\frac{1}{60.63}+\frac{1}{63.66}+\frac{1}{66.69}+...+\frac{1}{117.120}+\frac{1}{2011}\right)\)
\(=2\left(\frac{3}{60.63}+\frac{3}{63.66}+....+\frac{3}{117.120}+\frac{3}{2011}\right).\frac{1}{3}\)
\(=2\left(\frac{1}{60}-\frac{1}{63}+\frac{1}{63}-\frac{1}{66}+...+\frac{1}{117}-\frac{1}{120}+\frac{3}{2011}\right).\frac{1}{3}\)
\(=2\left(\frac{1}{60}-\frac{1}{120}+\frac{3}{2011}\right).\frac{1}{3}\)\(=\frac{2}{3}.\left(\frac{1}{120}+\frac{3}{2011}\right)=\frac{2}{3}.\frac{1}{120}+\frac{3}{2011}.\frac{2}{3}\)
\(=\frac{1}{180}+\frac{2}{2011}\)
B=\(5\left(\frac{1}{40.44}+\frac{1}{44.48}+...+\frac{1}{76.80}\right)+\frac{5}{2011}\)
\(=\frac{5}{4}\left(\frac{1}{40}-\frac{1}{44}+\frac{1}{44}-\frac{1}{48}+...+\frac{1}{76}-\frac{1}{80}\right)+\frac{5}{2011}\)
\(=\frac{5}{4}\left(\frac{1}{40}-\frac{1}{80}\right)+\frac{5}{2011}=\frac{5}{4}.\frac{1}{80}+\frac{5}{2011}\)\(=\frac{1}{64}+\frac{5}{2011}\)
Xét: \(\frac{1}{180}< \frac{1}{64};\frac{2}{2011}< \frac{5}{2011}\)
\(\Rightarrow\frac{1}{180}+\frac{2}{2011}< \frac{1}{64}+\frac{5}{2011}\)
\(\Leftrightarrow A< B\)
Vậy: A<B
9 tháng 5 2019
Bài 3: Ta có:
C=222...22000...00777....7
( có 2011 c/s 2; 2011 c/s 0; 2011 c/s 7)
\(\Rightarrow\) Tổng các c/s của C là:
2011.2+2011.0+2011.7=18099=9.2011 \(⋮9\)
\(\Rightarrow C⋮9\)
Vậy C có ít nhất 3 ước: 1;C và C.
Từ đó suy ra C là hợp số.
Vậy C là hợp số.
XO
11 tháng 10 2020
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
\(=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{5\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}-\frac{2\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}{3\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}\)
\(=\frac{1}{5}-\frac{2}{3}=-\frac{7}{15}\)
11 tháng 10 2020
Ta có:
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
\(P=\frac{1}{5}\cdot\left(\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}\right)-\frac{2}{3}\cdot\left(\frac{\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}}{\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}}\right)\)
\(P=\frac{1}{5}-\frac{2}{3}=-\frac{7}{15}\)
khó quá mấy anh chị ơi ! em mới học lớp 5 thôi .