\(x^3-y^3+xy=1\)

\(\Leftrightarrow\left(x-y\right)^3+3xy\left(x-y\right)+xy=1\)

\(\Leftrightarrow\left(x-y\right)^3+\frac{1}{27}+3xy\left(x-y+\frac{1}{3}\right)=\frac{26}{27}\)

\(\Leftrightarrow\left(x-y+\frac{1}{3}\right)\left[\left(x-y\right)^2-\frac{x-y}{3}+\frac{1}{9}\right]+3xy\left(x-y+\frac{1}{3}\right)=\frac{26}{27}\)

\(\left(x-y+\frac{1}{3}\right)\left[\left(x-y\right)^2-\frac{x-y}{3}+\frac{1}{9}+3xy\right]=\frac{26}{27}\) 

Đoạn này ez

\(\frac{x^3+y^3}{6}=\frac{x^3-2y^3}{4}=\frac{\left(x^3+y^3\right)-\left(x^3-2y^3\right)}{2}=\frac{3y^3}{2}\)

Từ\(\frac{x^3+y^3}{6}=\frac{3y^3}{2}\Rightarrow2x^3+2y^3=18y^3\Rightarrow2x^3=16y^3\Rightarrow x^3=8y^3=2^3y^3=\left(2y\right)^3\Rightarrow x=2y\)

Thế \(x=2y\)vào \(\left|xy\right|=\left|2y\cdot y\right|=2\Rightarrow\left|2y^2\right|=2\Rightarrow2y^2=2\)(vì \(2y^2\ge0\))\(\Rightarrow y^2=1\)

\(\Rightarrow y=\pm1\Rightarrow x=\pm2\)

có nghĩ là có 4 đáp số nhé bạn y=1;x=2

                                                 y=1;x=-2

                                                 y=-1;x=2

                                                 y=-1;x=-2

a) => 2xy +3x=y+1

=> 2xy+3x-y=1

=> x(2y+3) -  1/2 (2y+3) +3/2 =1

=> (x-1/2)(2y+3)=1-3/2= -1/2

=> (2x-1)(2y+3)=-1

ta có bảng

...........