\(S=-2,5+\frac{23}{4}-\left(-3,25+\frac{1}{2}+\frac{13}{2}\right)\)

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

\(S=-\frac{5}{2}+\frac{23}{4}-\left(\frac{15}{4}\right)\)

\(S=-\frac{1}{2}\)

\(S=-2,5+\frac{23}{4}-\left(-3,25+\frac{1}{2}+\frac{13}{2}\right)\)

\(S=-2,5+\frac{23}{4}+3,25-\frac{1}{2}-\frac{13}{2}\)

\(S=\frac{23}{4}-2,5+3,25-7\)

\(S=3,25+3,25-7\)

\(S=6,5-7\)

\(S=-0,5\)

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) \(2\frac{1}{4}:\left(\frac{-3}{5}\right)-1\frac{1}{4}:\left(\frac{-3}{5}\right)\)

\(=\frac{9}{4}.\left( \frac{-5}{3}\right)-\frac{5}{4}.\left(\frac{-5}{3}\right)\)

=\(\left(\frac{-5}{3}\right).\left(\frac{9}{4}-\frac{5}{4}\right)\)

\(=\frac{-5}{3}.1\)

\(=\frac{-5}{3}\)

Chữ hơi xấu tí

⚽

\(M=\frac{2.\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{23}\right)}{-5.\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{23}\right)}+\frac{\frac{1}{17}-\frac{1}{23}+\frac{1}{31}}{3.\left(\frac{1}{17}-\frac{1}{23}+\frac{1}{31}\right)}=-\frac{2}{5}+\frac{1}{3}=\frac{1}{15}.\)

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

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

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

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

Giúp mik với mn ơi

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. \(\frac{\frac{3}{7}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{7}-\frac{5}{11}+\frac{5}{13}}+\frac{0.5-\frac{1}{3}+\frac{1}{4}}{-\frac{3}{2}+1-\frac{3}{4}}\)

\(=\frac{3\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}{5\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{-3\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}\right)}\)

\(=\frac{3}{5}-\frac{1}{3}\)

\(=\frac{9}{13}-\frac{5}{15}=\frac{4}{15}\)

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

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