Question

\(\left(x-2\right)^2\times\left(y-3\right)=-4\)

Answer 1

Bạn có thể viết lại phép đó không

Answer 2

Ta có : (x - 2)² . (y - 3) = -4 = 1.(-4) = (-1).4 = 2.(-2) = (-2).2

Tiếp theo bạn kẻ bảng theo từng cặp là sẽ ra được các cặp x,y nha

Answer 3

\(\Rightarrow\left\{{}\begin{matrix}\left(x-2\right)^2=\left(-1\right)^2\\y-3=-4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left(x-2\right)=-1\\y=-4+3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1+2\\y=-1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x-2\right)^2=1^2\\y-3=-4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left(x-2\right)=1\\y=-4+3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1+2\\y=-1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x-2\right)^2=\left(-2\right)^2\\y-3=-1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left(x-2\right)=\left(-2\right)\\y=-1+3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-2+2\\y=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\y=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left(x-2\right)^2=2^2\\y-3=-1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-2=2\\y=-1+3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2+2\\y=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4\\y=2\end{matrix}\right.\)Vậy x=1,y=-1 ; x=3,y=-1 ; x=0,y=2 ; x=4,y=2