\(\dfrac{5}{2.7}+\dfrac{16}{7.9}-\dfrac{2}{9.11}-\dfrac{29}{11.18}\)

\(=\dfrac{5-2}{2.7}+\dfrac{7+9}{7.9}-\dfrac{11-9}{9.11}-\dfrac{11+18}{11.18}\)

\(=\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{1}{9}-\dfrac{1}{9}+\dfrac{1}{11}-\dfrac{1}{11}-\dfrac{1}{18}\)

\(=\dfrac{1}{2}-\dfrac{1}{18}=\dfrac{4}{9}\)

A = \(\dfrac{5}{2.7}\) + \(\dfrac{16}{7.9}\) - \(\dfrac{2}{9.11}\) - \(\dfrac{29}{11.18}\)

A = \(\dfrac{1}{2}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) + \(\dfrac{1}{9}\) - \(\dfrac{1}{9}\)  + \(\dfrac{1}{11}\) - \(\dfrac{1}{11}\) - \(\dfrac{1}{18}\)

A = \(\dfrac{1}{2}\) - \(\dfrac{1}{18}\)

A =  \(\dfrac{4}{9}\)

\(\dfrac{-3}{21}\) + \(\dfrac{6}{42}\) - \(\dfrac{-7}{49}\) = \(\dfrac{-1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{-1}{7}\) =  \(\dfrac{1}{7}\)

Lời giải:

\((1+\frac{1}{3})(1+\frac{1}{8})(1+\frac{1}{15})...(1+\frac{1}{99})\\ =\frac{4}{3}.\frac{9}{8}.\frac{16}{15}....\frac{100}{99}\\ =\frac{2^2.3^2.4^2....10^2}{1.3.2.4.3.5....9.11}=\frac{(2.3.4..10)(2.3.4..10)}{(1.2.3...9)(3.4.5...11)}=10.\frac{2}{11}=\frac{20}{11}\)

A = \(\dfrac{7}{18}\).[ - \(\dfrac{12}{23}\) - \(\dfrac{4}{15}\)] + \(\dfrac{7}{18}\).[\(\dfrac{8}{30}\)+ \(\dfrac{25}{23}\)]

A =- \(\dfrac{7}{18}\).\(\dfrac{12}{23}\) - \(\dfrac{7}{18}.\dfrac{4}{15}\) + \(\dfrac{7}{18}\).\(\dfrac{8}{30}\) + \(\dfrac{7}{18}\).\(\dfrac{25}{23}\)

A =  \(\dfrac{7}{18}\).(\(\dfrac{25}{23}\) - \(\dfrac{12}{23}\))  - (\(\dfrac{7}{18}\).\(\dfrac{4}{15}\) - \(\dfrac{7}{18}\).\(\dfrac{4}{15}\))

= \(\dfrac{7}{18}\).\(\dfrac{13}{23}\) - 0

= \(\dfrac{91}{414}\)

\(\dfrac{x}{25}=\dfrac{-3}{7}\cdot\dfrac{7}{6}\)

\(\Rightarrow\dfrac{x}{25}=\dfrac{-3\cdot7}{7\cdot6}\)

\(\Rightarrow\dfrac{x}{25}=\dfrac{-1}{2}\)

\(\Rightarrow x=-\dfrac{25}{2}\)

\(2x-y+4xy=6\)

\(\Rightarrow\left(4xy+2x\right)-y=6\)

\(\Rightarrow2x\left(2y+1\right)-y=6\)

\(\Rightarrow2\cdot2x\left(2y+1\right)-2y=2\cdot6\)

\(\Rightarrow4x\left(2y+1\right)-2y-1=11\)

\(\Rightarrow4x\left(2y+1\right)-\left(2y+1\right)=11\)

\(\Rightarrow\left(2y+1\right)\left(4x-1\right)=11\)

Mà x và y nguyên nên ta có bảng: 

Vậy các cặp (x;y) thỏa mãn là: \(\left(3;0\right);\left(0;-6\right)\)

Số gạo lúc sau ở mỗi thùng:

60 : 2 = 30 (kg)

Số gạo ở thùng thứ nhất lúc đầu:

30 : 3/4 = 40 (kg)

Số gạo ở thùng thứ hai lúc đầu:

60 - 40 = 20 (kg)

\(\left(1-\dfrac{2}{5}\right)\cdot\left(1-\dfrac{2}{7}\right)\cdot\left(1-\dfrac{2}{9}\right)\cdot...\cdot\left(1-\dfrac{2}{19}\right)\)

\(=\dfrac{5-2}{5}\cdot\dfrac{7-2}{7}\cdot\dfrac{9-2}{9}\cdot...\cdot\dfrac{19-2}{19}\)

\(=\dfrac{3}{5}\cdot\dfrac{5}{7}\cdot\dfrac{7}{9}\cdot...\cdot\dfrac{17}{19}\)

\(=\dfrac{3\cdot5\cdot7\cdot...\cdot17}{5\cdot7\cdot9\cdot...\cdot19}\)

\(=\dfrac{3}{19}\)

\(-\dfrac{1}{2}-\dfrac{2}{3}+\left(\dfrac{-3}{4}\right)+\dfrac{2}{3}\\ =-\dfrac{1}{2}-\dfrac{3}{4}-\left(\dfrac{2}{3}-\dfrac{2}{3}\right)\\ =-\dfrac{1}{2}-\dfrac{3}{4}=-\left(\dfrac{1}{2}+\dfrac{3}{4}\right)=-\dfrac{5}{4}\)

\(-\dfrac{4}{3}-\dfrac{3}{8}+\dfrac{7}{3}-\dfrac{3}{2}\\ =-\left(\dfrac{4}{3}-\dfrac{7}{3}\right)-\dfrac{3}{8}-\dfrac{3}{2}\\ =-\left(-\dfrac{3}{3}\right)-\dfrac{3}{8}-\dfrac{3}{2}\\ =1-\dfrac{3}{8}-\dfrac{3}{2}\\ =\dfrac{8}{8}-\dfrac{3}{8}-\dfrac{12}{8}=-\dfrac{7}{8}\)

b) \(-\dfrac{1}{2}-\dfrac{2}{3}+\dfrac{-3}{4}+\dfrac{2}{3}\)

\(=\left(-\dfrac{1}{2}+\dfrac{-3}{4}\right)+\left(-\dfrac{2}{3}+\dfrac{2}{3}\right)\)

\(=\left(\dfrac{-2}{4}+\dfrac{-3}{4}\right)+0\)

\(=\dfrac{-5}{4}\)

c) \(\dfrac{-4}{3}-\dfrac{3}{8}+\dfrac{7}{3}-\dfrac{3}{2}\) 

\(=\left(\dfrac{-4}{3}+\dfrac{7}{3}\right)+\left(-\dfrac{3}{8}-\dfrac{3}{2}\right)\)

\(=\dfrac{3}{3}+\left(-\dfrac{3}{8}-\dfrac{12}{8}\right)\)

\(=1-\dfrac{15}{8}\)

\(=-\dfrac{7}{8}\)

\(\left(-3\right)\cdot2^3+27:\left(-3\right)^2\)

\(=\left(-3\right)\cdot8+27:9\)

\(=-24+3\)

\(=-21\)

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