S = -2,5 + \(\frac{23}{4}\) - (-3,75 + \(\frac{1}{2}\) + \(\frac{13}{2}\) )
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\(S=-2,5+\frac{23}{4}-\left(-3,25+\frac{1}{2}+\frac{13}{2}\right)\)
\(S=-\frac{5}{2}+\frac{23}{4}-\left(-\frac{13}{4}+\frac{1}{2}+\frac{13}{2}\right)\)
\(S=-\frac{5}{2}+\frac{23}{4}-\left(\frac{15}{4}\right)\)
\(S=-\frac{1}{2}\)
\(S=-2,5+\frac{23}{4}-\left(-3,25+\frac{1}{2}+\frac{13}{2}\right)\)
\(S=-2,5+\frac{23}{4}+3,25-\frac{1}{2}-\frac{13}{2}\)
\(S=\frac{23}{4}-2,5+3,25-7\)
\(S=3,25+3,25-7\)
\(S=6,5-7\)
\(S=-0,5\)
\(M=\frac{2.\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{23}\right)}{-5.\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{23}\right)}+\frac{\frac{1}{17}-\frac{1}{23}+\frac{1}{31}}{3.\left(\frac{1}{17}-\frac{1}{23}+\frac{1}{31}\right)}=-\frac{2}{5}+\frac{1}{3}=\frac{1}{15}.\)
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
1.\(\left(-\frac{6}{5}+\frac{6}{16}-\frac{6}{23}\right):\left(\frac{9}{5}-\frac{9}{16}+\frac{9}{23}\right)\)
\(=6\left(-\frac{1}{5}+\frac{1}{16}-\frac{1}{23}\right):\left(-9\right)\left(\frac{-1}{5}+\frac{1}{16}-\frac{1}{23}\right)\)
\(=6:\left(-9\right)=-\frac{2}{3}\)
2. \(\frac{\frac{3}{7}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{7}-\frac{5}{11}+\frac{5}{13}}+\frac{0.5-\frac{1}{3}+\frac{1}{4}}{-\frac{3}{2}+1-\frac{3}{4}}\)
\(=\frac{3\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}{5\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{-3\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}\right)}\)
\(=\frac{3}{5}-\frac{1}{3}\)
\(=\frac{9}{13}-\frac{5}{15}=\frac{4}{15}\)
\(S=2,5+\frac{23}{4}-\left(-3,75+\frac{1}{2}+\frac{13}{2}\right)\)
\(S=\frac{-25}{10}+\frac{23}{4}+3,75+\frac{1}{2}+\frac{13}{2}\)
\(S=\frac{-5}{2}+\frac{23}{4}+\frac{-375}{100}+\frac{1}{2}+\frac{13}{2}\)
\(S=\frac{-5}{2}+\frac{23}{4}+\frac{-15}{4}+\frac{1}{2}+\frac{13}{2}\)
\(S=\frac{-10}{4}+\frac{23}{4}+\frac{-15}{4}+\frac{2}{4}+\frac{26}{4}\)
\(S=\frac{26}{4}=\frac{13}{2}\)
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