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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(=-24+3\)

\(=-21\)

-6

\(\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+...+\dfrac{1}{2015\cdot2018}\)

\(=\dfrac{1}{3}\cdot3\cdot\left(\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+...+\dfrac{1}{2015\cdot2018}\right)\)

\(=\dfrac{1}{3}\cdot\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{2015\cdot2018}\right)\)

\(=\dfrac{1}{3}\cdot\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{2015}-\dfrac{1}{2018}\right)\)

\(=\dfrac{1}{3}\cdot\left(\dfrac{1}{2}-\dfrac{1}{2018}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{504}{1009}\)

\(=\dfrac{168}{1009}\)

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