Đáp án:
\begin{array}{I}b)\\xy+3y-2y =11\\\to x(y+3)-2(y+3)+6=11\\\to(x-2)(y+3)\\\to x-2;y+3∈ Ư(5)=5\\\text{Ta có bảng sau}\end{array}
\begin{array}{|c|c|}\hline x-2&1&5&-1&-5\\\hline y+3&5&1&-5&-1\\\hline x&3&7&1&-3\\\hline y&2&-2&-8&-4\\\hline\end{array}
\begin{array}{I}\text{Vậy (x;y) ∈ {(3;2);(7;-2);(1;-8);(-3;-4)}}\end{array}
\begin{array}{I}\text{Bài 2}\\\text{Theo bài ra, áp dụng tính chất dãy tỉ số bằng nhau, ta có}\\\dfrac{4x-3y}{5}= \dfrac{5y-4z}{3}= \dfrac{3z-5x}{4} =\dfrac{20x-15y}{25}=\dfrac{15y-12z}{9}=\dfrac{12z-20z}{16}=\dfrac{20x-15y+15y-12z+12z-20x}{25}=\dfrac{0}{25+9+16}= \dfrac{0}{50}=0\\\to 4x-3y=0;5y-4z=0;3z-5x=0\\\to 4x=3y; 5y=4z; 3z=5x; \\\to \dfrac{x}{3}=\dfrac{y}{4}; \dfrac{y}{4} = \dfrac{z}{5}; \dfrac{x}{3} = \dfrac{z}{3}\\\to \dfrac{x}{3}= \dfrac{y}{4}=\dfrac{z}{5}=\dfrac{x-y-+z}{3-4+5}=\dfrac{2020}{4}= 505\\\to x = 505*3 = 1515 ; y = 505 *4= 2020; z = 505 *5 =2525 \\\text{Vậy x = 1515; y = 2020 ; z = 2525}\end{array}
