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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)\(\dfrac{29}{32}\left(\dfrac{41}{36}-\dfrac{32}{58}\right)-\dfrac{41}{36}\left(\dfrac{29}{32}+\dfrac{18}{41}\right)\)
=\(\dfrac{29}{32}.\dfrac{41}{36}-\dfrac{29}{32}.\dfrac{32}{58}-\dfrac{41}{36}.\dfrac{29}{32}+\dfrac{18}{41}.\dfrac{41}{36}\)
=\(\left(\dfrac{29}{32}.\dfrac{41}{36}-\dfrac{41}{36}\dfrac{29}{32}\right)-\dfrac{29}{32}.\dfrac{32}{58}+\dfrac{18}{41}.\dfrac{41}{36}\)
=\(0-\dfrac{1}{2}+\dfrac{1}{2}=0\)
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,\(\dfrac{1}{2}\).\(\dfrac{4}{3}\)-\(\dfrac{20}{3}\).\(\dfrac{4}{5}\)=\(\dfrac{2}{3}\)-\(\dfrac{16}{3}\)=-\(\dfrac{14}{3}\)
\(a.\)
\(\dfrac{17}{8}:\left(\dfrac{27}{8}+\dfrac{11}{2}\right)\)
\(=\dfrac{17}{8}:\left(\dfrac{27+44}{8}\right)=\dfrac{17}{8}:\dfrac{71}{8}=\dfrac{17}{8}\cdot\dfrac{8}{71}=\dfrac{17}{71}\)
\(b.\)
\(\dfrac{28}{15}\cdot\dfrac{1}{4^2}\cdot3+\left(\dfrac{8}{15}-\dfrac{69}{60}\cdot\dfrac{5}{23}\right):\dfrac{51}{54}\)
\(=\dfrac{28}{15}\cdot\dfrac{1}{4^2}\cdot3+\left(\dfrac{8}{15}-\dfrac{1}{4}\right):\dfrac{51}{54}\)
\(=\dfrac{28}{15}\cdot\dfrac{1}{4^2}\cdot3+\left(\dfrac{8\cdot4-15}{60}\right):\dfrac{51}{54}\)
\(=\dfrac{28}{15}\cdot\dfrac{1}{4^2}\cdot3+\dfrac{17}{60}:\dfrac{51}{54}\)
\(=\dfrac{28}{15}\cdot\dfrac{1}{16}\cdot3+\dfrac{17}{60}\cdot\dfrac{54}{51}\)
\(=\dfrac{7}{20}+\dfrac{3}{10}\)
\(=\dfrac{7+3\cdot2}{20}=\dfrac{13}{20}\)
\(=-\left[\dfrac{-17}{21}+\dfrac{-19}{51}-\dfrac{4}{21}\right]+\dfrac{32}{51}\)
\(=1+\dfrac{19}{51}+\dfrac{32}{51}=1+1=2\)
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!!!
mk có thắc mắc là bạn để 3 ra ngoài sao 1/3 vẫn giữ nguyên vậy phải bằng 1/9 mới đúng chứ'
\(\dfrac{23}{19}.\left(\dfrac{17}{29}-\dfrac{19}{23}\right)-\dfrac{17}{29}.\dfrac{4}{19}\)
\(=\dfrac{23}{19}.\dfrac{17}{29}-\dfrac{23}{19}.\dfrac{19}{23}-\dfrac{17}{29}.\dfrac{4}{19}\)
\(=\dfrac{23}{19}.\dfrac{17}{29}-\dfrac{17}{29}.\dfrac{4}{19}-1\)
\(=\dfrac{17}{19.29}.\left(23-4\right)-1\)
\(=\dfrac{17}{19.29}.19-1\)
\(=\dfrac{17}{29}-1\)
\(=-\dfrac{12}{29}\)
a)P=\(\dfrac{-2}{7}+\dfrac{14}{29}+\dfrac{12}{33}+\dfrac{15}{29}+\dfrac{21}{33}+\dfrac{2}{7}\)
=\(\left(\dfrac{-2}{7}+\dfrac{2}{7}\right)+\left(\dfrac{14}{29}+\dfrac{15}{29}\right)+\left(\dfrac{12}{33}+\dfrac{21}{33}\right)\)
=0+1+1=2
b)\(\dfrac{2}{7}.\dfrac{5}{19}+\dfrac{2}{7}.\dfrac{14}{19}+\dfrac{21}{19}-\dfrac{2}{7}.\dfrac{1}{5}\)
=\(\dfrac{2}{7}.\left(\dfrac{5}{19}+\dfrac{14}{19}-\dfrac{1}{5}\right)+\dfrac{21}{19}\)
=\(\dfrac{2}{7}.\dfrac{4}{5}+\dfrac{21}{19}=\dfrac{887}{665}\)
29/19.49/51+29/19.34/51-29/19.32/51
=29/19.(49/51+34/51-32/51)
=29/19.1
=29/19
\(\dfrac{29}{19}.\dfrac{49}{51}+\dfrac{29}{19}.\dfrac{34}{51}-\dfrac{29}{19}.\dfrac{32}{51}\)
\(=\dfrac{29}{19}.\left(\dfrac{49}{51}+\dfrac{34}{51}-\dfrac{32}{51}\right)\)
\(=\dfrac{29}{19}.1\)
\(=\dfrac{29}{19}\)