Question

Tìm x, y thuộc Z để:

a) xy + x - y = 2

b) x - 2xy + y = 0

c) x. (x - 2) - (2 - x)y - 2. (x - 2) = 3

d) (2x - y). (4x² + 2xy + y²) + (2x + y). (4x² - 2xy + y²) - 16x. (x² - y) = 32

e) x² - 2xy + 2y² - 2x + 6y +5 = 0

g) x² + 2xy + 7x + 7y + 2y² = 0

 

Answer 1

a) \(xy+x-y=2\)

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

\(\Leftrightarrow\left(x-1\right)\left(y+1\right)=1=1.1=\left(-1\right).\left(-1\right)\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=y+1=1\\x-1=y+1=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2;y=0\\x=0;y=-2\end{cases}}\)

b) \(x-2xy+y=0\)

\(\Leftrightarrow2x-4xy+2y=0\)

\(\Leftrightarrow2x\left(1-2y\right)-\left(1-2y\right)=-1\)

\(\Leftrightarrow\left(2x-1\right)\left(1-2y\right)=-1\)

Tương tự nha

Answer 2

c) \(x\left(x-2\right)-\left(2-x\right)y-2\left(x-2\right)=3\)

\(\Leftrightarrow x\left(x-2\right)+\left(x-2\right)y-2\left(x-2\right)=3\)

\(\Leftrightarrow\left(x-2\right)\left(x+y-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+y-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=0\end{cases}}\)