\(S=2,5+\frac{23}{4}-\left(-3,75+\frac{1}{2}+\frac{13}{2}\right)\)

\(S=\frac{-25}{10}+\frac{23}{4}+3,75+\frac{1}{2}+\frac{13}{2}\)

\(S=\frac{-5}{2}+\frac{23}{4}+\frac{-375}{100}+\frac{1}{2}+\frac{13}{2}\)

\(S=\frac{-5}{2}+\frac{23}{4}+\frac{-15}{4}+\frac{1}{2}+\frac{13}{2}\)

\(S=\frac{-10}{4}+\frac{23}{4}+\frac{-15}{4}+\frac{2}{4}+\frac{26}{4}\)

\(S=\frac{26}{4}=\frac{13}{2}\)

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

1. \(\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2003}+\dfrac{2}{2004}-\dfrac{2}{2005}}{\dfrac{3}{2003}+\dfrac{3}{2004}-\dfrac{3}{2005}}\)

=\(\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{5\cdot\left(\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}\right)}-\)\(\dfrac{2\cdot\left(\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}\right)}{3\cdot\left(\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}\right)}\)

=\(\dfrac{1}{5}-\dfrac{2}{3}\)

=\(-\dfrac{7}{15}\)

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!!!!!
 

\(-\frac{125}{668994}\)nha

cai nay ban cu tinh nhu binh thuong thoi

sai r nha bạn