Question

Tìm x; y thuộc Z:

a) ( x - 1)( x² + 1) = 0

b) xy + 3x - 2y = 11

Answer 1

a) \(\left(x-1\right)\left(x^2+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x^2+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x^2=-1\end{matrix}\right.\)

suy ra x=1 do \(x^2=-1\)ko có giá trị thỏa mãn

Answer 2

b, \(xy+3x-2y=11\)

\(\Rightarrow x\left(y+3\right)-2y-6=5\)

\(\Rightarrow x\left(y+3\right)-2\left(y+3\right)=5\)

\(\Rightarrow\left(x-2\right)\left(y+3\right)=5\)

Ta có bảng sau: \(\left(x;y\in Z\right)\)

\(x-2\)	1	-1	5	-5
\(y+3\)	5	-5	1	-1
x	3	1	7	-3
y	2	-8	-2	-4

Vậy...