7)\(\dfrac{-19}{34}\left(\dfrac{17}{19}+\dfrac{49}{18}\right)+\dfrac{49}{18}\left(\dfrac{19}{34}-\dfrac{18}{7}\right)\)

=\(\dfrac{-19}{34}.\dfrac{17}{19}+\dfrac{49}{18}.\dfrac{-19}{34}+\dfrac{49}{18}.\dfrac{19}{34}-\dfrac{18}{7}.\dfrac{49}{18}\)

=\(\dfrac{1}{2}+\left(\dfrac{49}{18}.\dfrac{-19}{34}+\dfrac{49}{18}.\dfrac{19}{34}\right)-7\)

=\(\dfrac{1}{2}+\left[\dfrac{49}{18}\left(\dfrac{-19}{34}+\dfrac{19}{34}\right)\right]-7\)

=\(\dfrac{1}{2}+0-7=\dfrac{-13}{2}\)

8)\(\dfrac{29}{32}\left(\dfrac{41}{36}-\dfrac{32}{58}\right)-\dfrac{41}{36}\left(\dfrac{29}{32}+\dfrac{18}{41}\right)\)

=\(\dfrac{29}{32}.\dfrac{41}{36}-\dfrac{29}{32}.\dfrac{32}{58}-\dfrac{41}{36}.\dfrac{29}{32}+\dfrac{18}{41}.\dfrac{41}{36}\)

=\(\left(\dfrac{29}{32}.\dfrac{41}{36}-\dfrac{41}{36}\dfrac{29}{32}\right)-\dfrac{29}{32}.\dfrac{32}{58}+\dfrac{18}{41}.\dfrac{41}{36}\)

=\(0-\dfrac{1}{2}+\dfrac{1}{2}=0\)

\(B=\dfrac{6}{19}\cdot\dfrac{-7}{11}+\dfrac{6}{19}\cdot\dfrac{-4}{11}+\dfrac{-13}{19}\)

\(B=\dfrac{6}{19}\cdot\left(-\dfrac{4}{11}+\dfrac{-7}{11}\right)+\dfrac{-13}{19}\)

\(B=\dfrac{6}{19}\cdot\left(-1\right)+\dfrac{-13}{19}\)

\(B=-\dfrac{6}{19}+\dfrac{-13}{19}=-\dfrac{19}{19}=-1\)

Vậy \(B=-1\)

\(A=\dfrac{\left(1+17\right).\left(1+\dfrac{17}{2}\right)..........\left(1+\dfrac{17}{19}\right)}{\left(1+19\right).\left(1+\dfrac{19}{2}\right)..........\left(1+\dfrac{19}{17}\right)}\)

\(=\dfrac{18.\dfrac{19}{2}.............\dfrac{36}{19}}{20.\dfrac{21}{2}..........\dfrac{36}{17}}\)

\(=\dfrac{18.19.20.......36}{1.2.3...19}:\dfrac{20.21.....36}{1.2.3...17}\)

\(=\dfrac{1.2.3......36}{1.2.....36}\)

\(=1\)

Bài 1:

\(a,=\frac{2}{3}-\frac{16}{3}=\frac{-14}{3}\)

\(b,=\left(\frac{3}{7}+\frac{4}{7}\right)+\left(-\frac{6}{19}+\frac{-13}{19}\right)=1-1=0\)

\(c,=\frac{3}{5}.\left(\frac{8}{9}-\frac{7}{9}+\frac{26}{9}\right)=\frac{3}{5}.3=\frac{9}{5}\)

a,\(\dfrac{1}{2}\).\(\dfrac{4}{3}\)-\(\dfrac{20}{3}\).\(\dfrac{4}{5}\)=\(\dfrac{2}{3}\)-\(\dfrac{16}{3}\)=-\(\dfrac{14}{3}\)

\(a.\)

\(\dfrac{17}{8}:\left(\dfrac{27}{8}+\dfrac{11}{2}\right)\)

\(=\dfrac{17}{8}:\left(\dfrac{27+44}{8}\right)=\dfrac{17}{8}:\dfrac{71}{8}=\dfrac{17}{8}\cdot\dfrac{8}{71}=\dfrac{17}{71}\)

\(b.\)

\(\dfrac{28}{15}\cdot\dfrac{1}{4^2}\cdot3+\left(\dfrac{8}{15}-\dfrac{69}{60}\cdot\dfrac{5}{23}\right):\dfrac{51}{54}\)

\(=\dfrac{28}{15}\cdot\dfrac{1}{4^2}\cdot3+\left(\dfrac{8}{15}-\dfrac{1}{4}\right):\dfrac{51}{54}\)

\(=\dfrac{28}{15}\cdot\dfrac{1}{4^2}\cdot3+\left(\dfrac{8\cdot4-15}{60}\right):\dfrac{51}{54}\)

\(=\dfrac{28}{15}\cdot\dfrac{1}{4^2}\cdot3+\dfrac{17}{60}:\dfrac{51}{54}\)

\(=\dfrac{28}{15}\cdot\dfrac{1}{16}\cdot3+\dfrac{17}{60}\cdot\dfrac{54}{51}\)

\(=\dfrac{7}{20}+\dfrac{3}{10}\)

\(=\dfrac{7+3\cdot2}{20}=\dfrac{13}{20}\)

Sai kìa phải bằng 11/20

\(=-\left[\dfrac{-17}{21}+\dfrac{-19}{51}-\dfrac{4}{21}\right]+\dfrac{32}{51}\)

\(=1+\dfrac{19}{51}+\dfrac{32}{51}=1+1=2\)

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!!!

a: \(=\dfrac{3}{8}\left(27+\dfrac{1}{5}-51-\dfrac{1}{5}\right)+19\)

\(=-24\cdot\dfrac{3}{8}+19=-9+19=10\)

b: \(=\left(35+\dfrac{1}{6}-46-\dfrac{1}{6}\right):\left(\dfrac{-4}{5}\right)\)

\(=\dfrac{-11\cdot5}{-4}=\dfrac{55}{4}\)

c: \(=\left(\dfrac{-15+8}{20}\right):\left[\dfrac{3}{7}+\dfrac{7}{3}\cdot\dfrac{12-5}{20}\right]\)

\(=\dfrac{-7}{20}:\left(\dfrac{3}{7}+\dfrac{49}{60}\right)\)

\(=-\dfrac{147}{523}\)