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\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
\(=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{5\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}-\frac{2\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}{3\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}\)
\(=\frac{1}{5}-\frac{2}{3}=-\frac{7}{15}\)
Ta có:
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
\(P=\frac{1}{5}\cdot\left(\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}\right)-\frac{2}{3}\cdot\left(\frac{\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}}{\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}}\right)\)
\(P=\frac{1}{5}-\frac{2}{3}=-\frac{7}{15}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{5\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}-\frac{2\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}{3\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}\)
\(P=\frac{1}{5}-\frac{2}{3}=\frac{3-10}{15}=\frac{-7}{15}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có \(2004+\frac{2003}{2}+\frac{2002}{3}+...+\frac{1}{2004}\)
\(=\left(1+\frac{2003}{2}\right)+\left(1+\frac{2002}{3}\right)...\left(1+\frac{1}{2004}\right)+1\)
\(=\frac{2005}{2}+\frac{2005}{3}+...+\frac{2005}{2004}+\frac{2005}{2005}\)
\(=2005\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2004}+\frac{1}{2005}\right)\)
\(\Rightarrow\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2005}}{\frac{2004}{1}+\frac{2003}{2}+\frac{2002}{3}+...+\frac{1}{2004}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2004}+\frac{1}{2005}}{2005\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2004}+\frac{1}{2005}\right)}\)
\(=\frac{1}{2005}\)
\(F=1-3+3^2-3^3+....+3^{2002}-3^{2003}+3^{2004}\)
\(3F=3-3^2+3^3-3^4+......+3^{2003}-3^{2004}+3^{2005}\)
\(3F+F=3^{2005}+1\)
\(\Leftrightarrow4F=3^{2005}+1\)
\(F=\frac{3^{2005}+1}{4}\)
Ta có: \(F=1-3+3^2-3^3+...+3^{2002}-3^{2003}+3^{2004}\)
\(\Rightarrow3F=3.\left(1-3+3^2-3^3+...+3^{2002}-3^{2003}+3^{2004}\right)\)
\(\Rightarrow3F=3-3^2+3^3-3^4+...+3^{2003}-3^{2004}+3^{2005}\)
\(\Rightarrow F+3F=1-3+3^2-3^3+...+3^{2002}-3^{2003}+3^{2004}+\)\(\left(3-3^2+3^3-3^4+...+3^{2003}-3^{2004}+3^{2005}\right)\)
\(\Rightarrow4F=1+3^{2005}\)
\(\Rightarrow F=\left(1+3^{2005}\right):4\)