1. Tính tổng: S= -2,5 +\(\frac{23}{4}\)-(-3,25+ \(\frac{1}{2}\)+ \(\frac{13}{2}\))
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\(S=2,5+\frac{23}{4}-\left(-3,75+\frac{1}{2}+\frac{13}{2}\right)\)
\(S=\frac{-25}{10}+\frac{23}{4}+3,75+\frac{1}{2}+\frac{13}{2}\)
\(S=\frac{-5}{2}+\frac{23}{4}+\frac{-375}{100}+\frac{1}{2}+\frac{13}{2}\)
\(S=\frac{-5}{2}+\frac{23}{4}+\frac{-15}{4}+\frac{1}{2}+\frac{13}{2}\)
\(S=\frac{-10}{4}+\frac{23}{4}+\frac{-15}{4}+\frac{2}{4}+\frac{26}{4}\)
\(S=\frac{26}{4}=\frac{13}{2}\)
a) ( -2.5 ) . ( 7,5) .( -4 )
= [(-2,5).(-4)].(7,5)
= 10 . 7,5
= 75
b) \(1\frac{4}{23}+\frac{8}{21}-\frac{4}{23}+0,6+\frac{13}{21}\)
=\(1\frac{4}{23}-\frac{4}{23}+\frac{8}{21}+\frac{13}{21}-0,6\)
\(=1+1-0,6\)
\(=2-0,6\)
= 1,4
c) \(\frac{2}{7}.15\frac{1}{3}-\frac{2}{7}.20.\frac{1}{3}+4\frac{1}{3}\)
\(=\frac{2}{7}.5-\frac{1}{3}.\frac{40}{7}+4\frac{1}{3}\)
= \(=\frac{10}{7}-\frac{17}{7}\)
= -1
d) \(2\frac{1}{4}:\left(\frac{-3}{5}\right)-1\frac{1}{4}:\left(\frac{-3}{5}\right)\)
\(=\frac{9}{4}.\left( \frac{-5}{3}\right)-\frac{5}{4}.\left(\frac{-5}{3}\right)\)
=\(\left(\frac{-5}{3}\right).\left(\frac{9}{4}-\frac{5}{4}\right)\)
\(=\frac{-5}{3}.1\)
\(=\frac{-5}{3}\)
\(=\dfrac{35.85\cdot14}{0.5+2.3}\cdot\dfrac{6}{17}\cdot\dfrac{16.8}{259,2}-\left(4.625-\dfrac{13}{6}:\dfrac{26}{3}\right):\left(3.25:2.25\right)\)
\(=\dfrac{1673}{408}-\dfrac{315}{104}=\dfrac{1421}{1326}\)
1 : 29 x ( 19 -13 ) - 19 x ( 29 - 13 )
= 29 x 6 - 19 x 16
= 174 - 304
= - 130
2 : 1 - \(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
= 1 - \(\frac{1}{100}\)
= \(\frac{99}{100}\)
Bài 1 :
\(S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2010}-\frac{1}{2011}\)
\(S=\frac{1}{1}-\frac{1}{2011}=\frac{2010}{2011}\)
Bài 2 :
\(S=\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}+...+\frac{1}{58}-\frac{1}{61}\)
\(S=\frac{1}{10}-\frac{1}{61}=\frac{51}{610}\)
Bài 3 :
\(3S=\frac{3}{4\times7}+\frac{3}{7\times11}+...+\frac{3}{19\times22}\)
\(3S=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{19}-\frac{1}{22}\)
\(3S=\frac{1}{4}-\frac{1}{22}\)
\(S=\frac{18}{88}\div3=\frac{6}{88}\)
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
a) \(\frac{1}{9}+3,25+5\frac{3}{16}+4\frac{1}{3}+2,8+0,5=\frac{1}{9}+\frac{13}{4}+\frac{83}{16}+\frac{13}{3}+\frac{14}{5}+\frac{1}{2}\)
\(=\frac{11651}{720}\)
B) \(2\frac{1}{3}+0,45+4,25+\frac{1}{81}+6\frac{8}{27}=\frac{7}{3}+\frac{9}{20}+\frac{17}{4}+\frac{1}{81}+\frac{170}{27}\)
\(=\frac{10807}{810}\)
C) \(1,25+2\frac{1}{4}+4\frac{2}{5}+0,3+2,14+4\frac{1}{8}=\frac{5}{4}+\frac{9}{4}+\frac{22}{5}+\frac{3}{10}+\frac{107}{50}+\frac{33}{8}\)
\(=\frac{2893}{200}\)
CHÚC BN HỌC TỐT!!!!!
\(S=-2,5+\frac{23}{4}-\left(-3,25+\frac{1}{2}+\frac{13}{2}\right)\)
\(S=-\frac{5}{2}+\frac{23}{4}-\left(-\frac{13}{4}+\frac{1}{2}+\frac{13}{2}\right)\)
\(S=-\frac{5}{2}+\frac{23}{4}-\left(\frac{15}{4}\right)\)
\(S=-\frac{1}{2}\)
\(S=-2,5+\frac{23}{4}-\left(-3,25+\frac{1}{2}+\frac{13}{2}\right)\)
\(S=-2,5+\frac{23}{4}+3,25-\frac{1}{2}-\frac{13}{2}\)
\(S=\frac{23}{4}-2,5+3,25-7\)
\(S=3,25+3,25-7\)
\(S=6,5-7\)
\(S=-0,5\)