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a) \(\frac{31}{17}+\frac{-5}{13}+\frac{-8}{13}-\frac{14}{17}\)
\(=\left(\frac{31}{17}-\frac{14}{17}\right)+\left(\frac{-5}{13}+\frac{-8}{13}\right)\)
\(=1+\left(-1\right)\)
\(=0\)
b) \(\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}+\frac{5}{7}\)
\(=\frac{-5}{7}.\left(\frac{2}{11}+\frac{9}{11}\right)+\frac{5}{7}\)
\(=\frac{-5}{7}.1+\frac{5}{7}\)
\(=\frac{-5}{7}+\frac{5}{7}\)
\(=0\)
Ta có:
\(\frac{A}{5}=\frac{4}{35\cdot31}+\frac{6}{35\cdot41}+\frac{9}{50\cdot41}+\frac{7}{50\cdot57}=\frac{35-31}{35\cdot31}+\frac{41-35}{35\cdot41}+\frac{50-41}{50\cdot41}+\frac{57-50}{50\cdot57}\) ( Ps cuối là\(\frac{57-50}{50.57}\) nha).
\(=\frac{1}{31}-\frac{1}{35}+\frac{1}{35}-\frac{1}{41}+\frac{1}{41}-\frac{1}{50}+\frac{1}{50}-\frac{1}{57}\)
\(\Rightarrow\frac{A}{5}=5\cdot\left(\frac{1}{31}-\frac{1}{57}\right)\)
Tương tự:
\(\frac{B}{2}=\frac{7}{38.31}+\frac{5}{38.43}+\frac{3}{43.46}+\frac{11}{46.57}=\frac{38-31}{38.31}+\frac{41-38}{38.41}+\frac{46-43}{43.46}+\frac{57-46}{46.57}\)
\(=\frac{1}{31}-\frac{1}{38}+\frac{1}{38}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}+\frac{1}{46}-\frac{1}{57}\)
\(\Rightarrow\frac{B}{2}=2\cdot\left(\frac{1}{31}-\frac{1}{57}\right)\)
Từ đó, suy ra:
\(\frac{A}{B}=\frac{5}{2}\)
Vậy \(\frac{A}{B}=\frac{5}{2}\)
\(\frac{5}{11}< \frac{a}{b}< \frac{5}{9}\)
=> \(\frac{45}{99}< \frac{a}{b}< \frac{55}{99}\)
=> b = 99 ; a = 46 -> 54