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\(=\dfrac{\dfrac{1}{2}\left(\dfrac{1}{7}-\dfrac{1}{15}-\dfrac{1}{23}\right)}{\dfrac{2}{5}\left(\dfrac{1}{7}-\dfrac{1}{15}-\dfrac{1}{23}\right)}:\left(\dfrac{2621}{4216}:\dfrac{647}{3162}\right)\)
\(=\dfrac{5}{4}:\dfrac{7863}{2588}=\dfrac{3235}{7863}\)
\(A=17\dfrac{2}{31}-\left(\dfrac{15}{17}+6\dfrac{2}{31}\right)=17\dfrac{2}{31}-\dfrac{15}{17}-6\dfrac{2}{31}\)
\(=11-\dfrac{15}{17}=\dfrac{172}{17}\)
\(B=\left(31\dfrac{6}{13}+5\dfrac{9}{41}\right)-36\dfrac{6}{12}=36\dfrac{363}{533}-36\dfrac{6}{12}=\dfrac{193}{1066}\)
\(C=27\dfrac{51}{59}-\left(7\dfrac{51}{59}-\dfrac{1}{3}\right)=27\dfrac{51}{59}-7\dfrac{51}{59}+\dfrac{1}{3}=20+\dfrac{1}{3}=\dfrac{61}{3}\)
\(A=17\dfrac{2}{31}-\left(\dfrac{15}{17}+6\dfrac{2}{31}\right)=17\dfrac{2}{31}-\dfrac{15}{17}-6\dfrac{2}{31}\)
\(=\left(17\dfrac{2}{31}-6\dfrac{2}{31}\right)-\dfrac{15}{17}=11-\dfrac{15}{17}=\dfrac{172}{17}\)
\(B=\left(31\dfrac{6}{13}+5\dfrac{9}{41}\right)-36\dfrac{6}{12}=36\dfrac{363}{533}-36\dfrac{1}{2}=\dfrac{193}{1066}\) (Casio :>)
\(C=27\dfrac{51}{59}-\left(7\dfrac{51}{59}-\dfrac{1}{3}\right)=27\dfrac{51}{59}-7\dfrac{51}{59}+\dfrac{1}{3}\)
\(=20+\dfrac{1}{3}=\dfrac{61}{3}\)
Bài 1:
a: \(=17+\dfrac{2}{31}-\dfrac{15}{17}-6-\dfrac{2}{31}=11-\dfrac{15}{17}=\dfrac{172}{17}\)
b: \(=31+\dfrac{6}{13}+5+\dfrac{9}{41}-36-\dfrac{9}{41}-36-\dfrac{6}{13}\)
=36
c: \(=27+\dfrac{51}{59}-7-\dfrac{51}{59}+\dfrac{1}{3}=20+\dfrac{1}{3}=\dfrac{61}{3}\)
Ta có:
\(A=\frac{3}{1\cdot5}+\frac{3}{5\cdot10}+...+\frac{3}{100\cdot105}\)
\(=\frac{3}{5}\cdot\left(\frac{5}{1\cdot5}+\frac{5}{5\cdot10}+...+\frac{5}{100\cdot105}\right)\)
\(=\frac{3}{5}\cdot\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{105}\right)\)
\(=\frac{3}{5}\left(1-\frac{1}{105}\right)=\frac{3}{5}\cdot\frac{104}{105}=\frac{312}{525}\)
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: \(=\left(\dfrac{-15+8}{20}\right):\left[\dfrac{3}{7}+\dfrac{7}{3}\cdot\dfrac{12-5}{20}\right]\)
\(=\dfrac{-7}{20}:\left(\dfrac{3}{7}+\dfrac{49}{60}\right)\)
\(=-\dfrac{147}{523}\)