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1: \(=\dfrac{16}{15}\left(-\dfrac{4}{9}+\dfrac{3}{7}\right)+\dfrac{16}{15}\left(\dfrac{4}{7}-\dfrac{5}{9}\right)\)
\(=\dfrac{16}{15}\left(-\dfrac{4}{9}+\dfrac{3}{7}+\dfrac{4}{7}-\dfrac{5}{9}\right)=0\)
2: \(=\dfrac{29}{9}\left(15+\dfrac{4}{7}-8-\dfrac{1}{7}+\dfrac{15}{7}-\dfrac{1}{7}\right)\)
\(=\dfrac{20}{9}\cdot\left(7\cdot\dfrac{18}{7}\right)=\dfrac{20}{9}\cdot18=40\)
\(A=\frac{1}{1\cdot2}+\frac{2}{2\cdot4}+\frac{3}{4\cdot7}+\frac{4}{7\cdot11}+...+\frac{10}{46\cdot56}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{46}-\frac{1}{56}\)
\(A=1-\frac{1}{56}\)
\(A=\frac{55}{56}\)
\(B=\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{23\cdot27}\)
\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{23}-\frac{1}{27}\)
\(B=\frac{1}{3}-\frac{1}{27}\)
\(B=\frac{8}{27}\)
\(C=\frac{4}{3\cdot6}+\frac{4}{6\cdot9}+\frac{4}{9\cdot12}+...+\frac{4}{99\cdot102}\)
\(C=\frac{4}{3}\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+\frac{3}{9\cdot12}+...+\frac{3}{99\cdot102}\right)\)
\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{99}-\frac{1}{102}\right)\)
\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{102}\right)\)
\(C=\frac{4}{3}\cdot\frac{33}{102}\)
\(C=\frac{22}{51}\)
\(a,\dfrac{9}{17}+\dfrac{15}{4}=\dfrac{36}{68}+\dfrac{255}{68}=\dfrac{291}{68}\\ b,\dfrac{19}{121}-\dfrac{1}{11}=\dfrac{19}{121}-\dfrac{11}{121}=\dfrac{8}{121}\\ c,\dfrac{4}{15}+3=\dfrac{4}{15}+\dfrac{45}{15}=\dfrac{49}{15}\\ d,3-\dfrac{5}{9}=\dfrac{27}{9}-\dfrac{5}{9}=\dfrac{22}{9}\)
\(a,\dfrac{9}{17}+\dfrac{15}{4}=\dfrac{36}{68}+\dfrac{255}{68}=\dfrac{291}{68}\)
\(b,\dfrac{19}{121}-\dfrac{1}{11}=\dfrac{19}{121}-\dfrac{11}{121}=\dfrac{8}{121}\)
\(c,\dfrac{4}{15}+3=\dfrac{4}{15}+\dfrac{3}{1}=\dfrac{4}{15}+\dfrac{45}{15}=\dfrac{49}{15}\)
\(d,3-\dfrac{5}{9}=\dfrac{3}{1}-\dfrac{5}{9}=\dfrac{27}{9}-\dfrac{5}{9}=\dfrac{22}{9}\)
1) Ta có: \(\frac{-4}{7}-\frac{11}{19}+\frac{13}{19}\cdot\frac{-3}{7}+\frac{2}{19}:\frac{-7}{4}\)
\(=\frac{-4}{7}-\frac{11}{19}-\frac{39}{133}-\frac{8}{133}\)
\(=\frac{-76}{133}-\frac{77}{133}-\frac{39}{133}-\frac{8}{133}\)
\(=\frac{-200}{133}\)
2) Ta có: \(\left(\frac{-4}{9}+\frac{3}{5}\right):\frac{1}{\frac{1}{5}}+\left(\frac{1}{5}-\frac{5}{9}\right):\frac{1}{\frac{1}{5}}\)
\(=\left(\frac{-4}{9}+\frac{3}{5}\right)\cdot\frac{1}{5}+\left(\frac{1}{5}-\frac{5}{9}\right)\cdot\frac{1}{5}\)
\(=\frac{1}{5}\left(\frac{-4}{9}+\frac{3}{5}+\frac{1}{5}-\frac{5}{9}\right)\)
\(=\frac{1}{5}\left(-1+\frac{4}{5}\right)\)
\(=\frac{1}{5}\cdot\frac{-1}{5}=\frac{-1}{25}\)
3) Ta có: \(\frac{4}{5}-\left(-\frac{2}{7}\right)-\frac{7}{10}\)
\(=\frac{4}{5}+\frac{2}{7}-\frac{7}{10}\)
\(=\frac{56}{70}+\frac{20}{70}-\frac{49}{70}\)
\(=\frac{27}{70}\)
4) Ta có: \(\frac{2}{7}-\left(-\frac{13}{15}+\frac{4}{9}\right)-\left(\frac{5}{9}-\frac{2}{15}\right)\)
\(=\frac{2}{7}+\frac{13}{15}-\frac{4}{9}-\frac{5}{9}+\frac{2}{15}\)
\(=\frac{2}{7}+1-1=\frac{2}{7}\)
\(\dfrac{3}{9}+\dfrac{9}{15}+\dfrac{16}{24}=\dfrac{1}{3}+\dfrac{9}{15}+\dfrac{2}{3}=\dfrac{5}{15}+\dfrac{9}{15}+\dfrac{10}{15}=\dfrac{24}{15}=\dfrac{8}{5}\)
\(\dfrac{4}{10}+\dfrac{12}{32}+\dfrac{9}{15}=\dfrac{2}{5}+\dfrac{3}{8}+\dfrac{3}{5}=\dfrac{2}{5}+\dfrac{3}{5}+\dfrac{3}{8}=\dfrac{5}{5}+\dfrac{3}{8}=1+\dfrac{3}{8}=1\dfrac{3}{8}=\dfrac{11}{8}\)
\(\dfrac{1}{6}+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{18}=\dfrac{3}{18}+\dfrac{9}{18}+\dfrac{6}{18}+\dfrac{1}{18}=\dfrac{19}{18}\)
\(\frac{4}{15}\)\(\div\)\(\frac{1}{3}\)\(+\)\(\frac{1}{15}\)\(\div\)\(\frac{1}{3}\)\(+\)\(\frac{4}{9}\)
\(=\)\(\frac{4}{15}\)\(\times\)\(3\)\(+\)\(\frac{1}{15}\)\(\times\)\(3\)\(+\)\(\frac{4}{9}\)
\(=\)\(3\)\(\times\)\(\left(\frac{4}{15}+\frac{1}{15}\right)\)\(+\)\(\frac{4}{9}\)
\(=\)\(3\)\(\times\)\(\frac{5}{15}\)\(+\)\(\frac{4}{9}\)
\(=\)\(3\)\(\times\)\(\frac{1}{3}\)\(+\)\(\frac{4}{9}\)
\(=\)\(1\)\(+\)\(\frac{4}{9}\)
\(=\)\(\frac{13}{9}\)
\(\frac{4}{15}\div\frac{1}{3}+\frac{1}{15}\div\frac{1}{3}+\frac{4}{9}\)
\(=\left(\frac{4}{15}+\frac{1}{15}\right)\times3+\frac{4}{9}\)
\(=\frac{1}{3}\times3+\frac{4}{9}\)
\(=1+\frac{4}{9}=\frac{13}{9}\)