\(A=\dfrac{636363\cdot37-373737\cdot63}{1+2+3+...+2006}\)

\(=\dfrac{37^2\cdot3^3\cdot7^2\cdot13-37^2\cdot3^3\cdot7^2\cdot13}{\left(2006+1\right)\cdot1003}\)

=0

S = ???

2

2

Ta có:B=1\(\dfrac{6}{41}\)( \(\dfrac{12+\dfrac{12}{19}-\dfrac{12}{37}-\dfrac{12}{53}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2006}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2006}}\) )

B=\(\dfrac{47}{41}\) [\(\dfrac{12\left(1+\dfrac{1}{19}-\dfrac{1}{37}-\dfrac{1}{53}\right)}{3\left(1+\dfrac{1}{3}-\dfrac{1}{37}-\dfrac{1}{53}\right)}:\dfrac{4\left(\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2006}\right)}{5\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2006}\right)}\) B = \(\dfrac{47}{41}\) [ \(\dfrac{12}{3}:\dfrac{4}{5}\)]

B = \(\dfrac{47}{41}\)[ 4 . \(\dfrac{5}{4}\)]

B = \(\dfrac{47}{41}.5\)

B = \(\dfrac{235}{41}\)

Chúc bn hc tốt!!!

mk có thắc mắc là bạn để 3 ra ngoài sao 1/3 vẫn giữ nguyên vậy phải bằng 1/9 mới  đúng chứ'

tìm 1 cái bằng 0

a: \(=\dfrac{-12}{7}\left(\dfrac{4}{35}+\dfrac{31}{35}\right)-\dfrac{2}{7}=\dfrac{-12}{7}-\dfrac{2}{7}=-2\)

b: =(-4)+(-4)+...+(-4)

=-4*25=-100

c: \(=157\cdot\left(-37\right)-41\cdot53+37\cdot157+51\cdot53\)

=10*53

=530

\(-1\dfrac{1}{5}.\dfrac{12+\dfrac{4}{3}-\dfrac{12}{37}-\dfrac{12}{35}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{35}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2003}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2003}}\)

\(=\dfrac{-6}{5}.\dfrac{4\left(3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{35}\right)}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{35}}:\dfrac{4\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}{5\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}\)

\(=\dfrac{-6}{5}.4:\dfrac{4}{5}\)

\(=\dfrac{-6.4.5}{5.4}=-6\)

= -4 đúng không

\(-1\dfrac{1}{5}.\dfrac{12+\dfrac{4}{3}-\dfrac{12}{37}-\dfrac{12}{35}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2003}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2003}}\)

\(=\dfrac{-6}{5}.\dfrac{4\left(3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}\right)}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}{5\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}\)

\(=\dfrac{-6}{5}.\dfrac{4}{3}:\dfrac{4}{5}\)

\(=\dfrac{-6.4.5}{5.3.4}=\dfrac{-6}{3}=-2\)

Vậy...

`8/7+419-1/7`

`=7/7+419`

`=1+419=420`

`-4/41+(-37/41+2)`

`=-(4/41+37/41)+2`

`=-1+2`

`=1`

\(\dfrac{8}{7}+419+\dfrac{-1}{7}\) 

=\(\left(\dfrac{8}{7}+\dfrac{-1}{7}\right)+419\) 

=1+419

=420

\(\dfrac{-4}{41}+\left(\dfrac{-37}{41}+2\right)\) 

=\(\dfrac{-4}{41}+\dfrac{-37}{41}+2\) 

=\(\left(\dfrac{-4}{41}+\dfrac{-37}{41}\right)+2\) 

=-1+2

=1

a: =>3/5x=5/6+3/2=5/6+9/6=14/6=7/3

=>x=35/9

b: =(4/5+16/5)+(37/7+57)

=61+37/7=464/7

\(\dfrac{3}{5}x=\dfrac{5}{6}+\dfrac{3}{2}=\dfrac{14}{6}=\dfrac{7}{3}\Leftrightarrow x=\dfrac{7}{3}:\dfrac{3}{5}=\dfrac{35}{9}\)

\(\dfrac{19}{5}+\dfrac{37}{7}+57=\dfrac{19.7+37.5}{35}+\dfrac{57.35}{35}=\dfrac{2313}{35}\)

\(\dfrac{3}{5}\times x-\dfrac{3}{2}=\dfrac{5}{6}\)

\(\dfrac{3}{5}\times x=\dfrac{5}{6}+\dfrac{3}{2}\)

\(\dfrac{3}{5}\times x=\dfrac{7}{3}\)

\(x=\dfrac{7}{3}:\dfrac{3}{5}\)

\(x=\dfrac{7}{3}\times\dfrac{5}{3}\)

\(x=\dfrac{35}{9}\)

\(\dfrac{12}{15}+\dfrac{37}{7}+\dfrac{16}{5}+57=\dfrac{4}{5}+\dfrac{37}{3}+\dfrac{16}{5}+57=\left(\dfrac{4}{5}+\dfrac{16}{5}\right)+\left(\dfrac{37}{7}+57\right)=4+\dfrac{436}{7}=\dfrac{464}{7}\)

a: \(=\dfrac{37}{4}+\dfrac{117}{16}+\dfrac{1}{4}=\dfrac{19}{2}+\dfrac{117}{16}=\dfrac{269}{16}\)

b: \(=1+\left(\dfrac{9}{10}+\dfrac{8}{10}\right):\dfrac{19}{6}=1+\dfrac{17}{10}\cdot\dfrac{6}{19}=\dfrac{146}{95}\)

c: \(=\dfrac{1}{4}-\dfrac{6}{4}+\dfrac{6}{5}=\dfrac{-5}{4}+\dfrac{6}{5}=\dfrac{-1}{20}\)