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\(A=29\dfrac{1}{2}\cdot\dfrac{2}{3}+39\dfrac{1}{3}\cdot\dfrac{3}{4}+\dfrac{5}{6}\)
\(=\dfrac{59}{2}\cdot\dfrac{2}{3}+\dfrac{118}{3}\cdot\dfrac{3}{4}+\dfrac{5}{6}\)
\(=\dfrac{59}{3}+\dfrac{118}{4}+\dfrac{5}{6}\)
\(=\dfrac{59}{3}+\dfrac{59}{2}+\dfrac{5}{6}\)
\(=59\cdot\left(\dfrac{1}{3}+\dfrac{1}{2}\right)+\left(\dfrac{1}{3}+\dfrac{1}{2}\right)\)
\(=\dfrac{5}{6}\cdot\left(59+1\right)=\dfrac{5}{6}\cdot60=50\)
2020/2019 x 2019/2018 x 2018/2017 x....................3/2
= 2020/2
= 1010
a) \(M=\frac{2\times2}{1\times5}+\frac{2\times2}{5\times9}+\frac{2\times2}{9\times13}+...+\frac{2\times2}{45\times40}\)
\(M=\frac{4}{1\times5}+\frac{4}{5\times9}+\frac{4}{9\times13}+...+\frac{4}{45\times49}\)
\(M=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{45}-\frac{1}{49}\)
\(M=1-\frac{1}{49}\)
\(M=\frac{48}{49}\)
b) \(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+5+...+10}\)
= \(\frac{2}{2\times\left(1+2\right)}+\frac{2}{2\times\left(1+2+3\right)}+...+\frac{2}{2\times\left(1+2+3+...+10\right)}\)
\(=\frac{2}{6}+\frac{2}{12}+...+\frac{2}{110}\)
\(=\frac{2}{2\times3}+\frac{2}{3\times4}+...+\frac{2}{10\times11}\)
\(=2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(=2\times\left(\frac{1}{2}-\frac{1}{11}\right)\)
\(=2\times\frac{9}{22}\)
\(=\frac{9}{11}\)
Mình trả lời câu a nha M= 4/1*5+4/5*9+4/9*13+...+4/45*49 M=1-1/5+1/5-1/9+1/9-1/13+...+1/45-1/49 M=1-1/49=48/49
