\(=\dfrac{3.4-\dfrac{3.4}{4}-\dfrac{3.4}{289}-\dfrac{3.4}{85}}{4-\dfrac{4}{7}-\dfrac{4}{289}-\dfrac{4}{85}}=\dfrac{3.\left(4-\dfrac{4}{7}-\dfrac{4}{289}-\dfrac{4}{85}\right)}{1.\left(4-\dfrac{4}{7}-\dfrac{4}{289}-\dfrac{4}{85}\right)}=\dfrac{3}{1}=3\)

=3

=3

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

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

=0

bạn ơi 4, là j đấy

`4/15:12/15+5/12:15/11+7/12`

`=4/12+11/36+7/12`

`=12/36+11/36+21/36`

`=44/36`

`=11/9`

1: Ta có: \(\dfrac{x+4}{4}+\dfrac{3x-7}{5}=\dfrac{7x+2}{20}\)

\(\Leftrightarrow5x+20+12x-28=7x+2\)

\(\Leftrightarrow17x-7x=2+8=10\)

hay x=1

2: Ta có: \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)

\(\Leftrightarrow\dfrac{6x}{36}+\dfrac{4\left(1-3x\right)}{36}=\dfrac{3\left(-x+1\right)}{36}\)

\(\Leftrightarrow6x+4-12x=-3x+3\)

\(\Leftrightarrow-6x+3x=3-4\)

hay \(x=\dfrac{1}{3}\)

3: Ta có: \(\dfrac{x-3}{3}-\dfrac{x+2}{12}=\dfrac{2x-1}{4}\)

\(\Leftrightarrow4x-12-x-2=6x-3\)

\(\Leftrightarrow3x-14-6x+3=0\)

\(\Leftrightarrow-3x=11\)

hay \(x=-\dfrac{11}{3}\)

4: Ta có: \(\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}\)

\(\Leftrightarrow3x-6-8x-12=x+6\)

\(\Leftrightarrow-5x-x=6+18\)

hay x=-4

5: Ta có: \(\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}\)

\(\Leftrightarrow6x-3+2x-6=-1\)

\(\Leftrightarrow8x=8\)

hay x=1

4=48/12

5/6=10/12

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

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

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

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

= -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...

\(=\dfrac{4}{11}\left(\dfrac{5}{12}-1+\dfrac{7}{12}\right)=\dfrac{4}{11}\cdot0=0\)

0

\(\dfrac{5}{12}.\dfrac{4}{11}-\dfrac{4}{11}+\dfrac{7}{12}.\dfrac{4}{11}\)

\(=\dfrac{4}{11}.\left(\dfrac{5}{12}-1+\dfrac{7}{12}\right)\)

\(=\dfrac{4}{11}.\left(\dfrac{5}{12}-\dfrac{12}{12}+\dfrac{7}{12}\right)\)

\(=\dfrac{4}{11}.\left(\dfrac{-7}{12}+\dfrac{7}{12}\right)\)

\(=\dfrac{4}{11}.\dfrac{0}{12}\)

\(=\dfrac{4}{11}.0\)

\(=0\)

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!!!

\(\dfrac{-1}{7}-\dfrac{1}{8}=\dfrac{-8}{56}-\dfrac{7}{56}=\dfrac{-15}{56}\\ \dfrac{-15}{48}-\dfrac{1}{12}=\dfrac{-5}{16}-\dfrac{1}{12}=\dfrac{-15}{48}-\dfrac{4}{48}=\dfrac{-19}{48}\\ \dfrac{-3}{4}-\dfrac{4}{5}=\dfrac{-15}{20}-\dfrac{16}{20}=\dfrac{-31}{20}\\ \dfrac{3}{4}+\dfrac{-5}{6}-\dfrac{11}{12}=\dfrac{9}{12}-\dfrac{10}{12}-\dfrac{11}{12}=\dfrac{-12}{12}=-1\)

\(=81.\dfrac{12.\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}{4.\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}:\dfrac{5.\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}{6.\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}.\dfrac{158}{711}\)

\(=81.\dfrac{12}{4}:\dfrac{5}{6}.\dfrac{2}{9}\)

\(=243:\dfrac{5}{6}.\dfrac{2}{9}\)

\(=\dfrac{1458}{5}.\dfrac{2}{9}\)

\(=\dfrac{324}{5}\)