Question

3)tim x

a)(x-1).(y-2)=5

b)x.(y-3)=12

Answer 1

a/ Ta có :

\(\left(x-1\right)\left(y-2\right)=5\)

Vì \(x,y\in N\Leftrightarrow x-1;y-2\in N\)\(,x-1;y-2\inƯ\left(5\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1=1\\y-2=5\end{matrix}\right.\\\left\{{}\begin{matrix}x-1=5\\y-2=1\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2\\y=7\end{matrix}\right.\\\left\{{}\begin{matrix}x=6\\y=3\end{matrix}\right.\end{matrix}\right.\)

Vậy ...........

b/ tương tự

Answer 2

a ) \(\left(x-1\right)\left(y-2\right)=5\)

Xảy ra 4 TH :

TH1 : \(\left[{}\begin{matrix}x-1=5\\y-2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\y=3\end{matrix}\right.\)

TH2 : \(\left[{}\begin{matrix}x-1=-5\\y-2=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\y=1\end{matrix}\right.\)

TH3 : \(\left[{}\begin{matrix}x-1=1\\y-2=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\y=7\end{matrix}\right.\)

TH4 : \(\left[{}\begin{matrix}x-1=-1\\y-2=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\y=-3\end{matrix}\right.\)

Vậy ........

b ) \(x\left(y-3\right)=12\)

Có 12TH xảy ra :

TH1 : \(\left[{}\begin{matrix}x=1\\y-3=12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\y=15\end{matrix}\right.\)

TH2 : \(\left[{}\begin{matrix}x=-1\\y-3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\y=-9\end{matrix}\right.\)

TH3 : \(\left[{}\begin{matrix}x=12\\y-3=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\y=4\end{matrix}\right.\)

TH4 : \(\left[{}\begin{matrix}x=-12\\y-3=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-12\\y=2\end{matrix}\right.\)

TH5 : \(\left[{}\begin{matrix}x=2\\y-3=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\y=9\end{matrix}\right.\)

TH6 : \(\left[{}\begin{matrix}x=-2\\y-3=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\y=-3\end{matrix}\right.\)

TH7 : \(\left[{}\begin{matrix}x=6\\y-3=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\y=5\end{matrix}\right.\)

TH8 : \(\left[{}\begin{matrix}x=-6\\y-3=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\y=1\end{matrix}\right.\)

TH9 : \(\left[{}\begin{matrix}x=3\\y-3=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\y=7\end{matrix}\right.\)

TH10 : \(\left[{}\begin{matrix}x=-3\\y-3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\y=-1\end{matrix}\right.\)

TH11 : \(\left[{}\begin{matrix}x=4\\y-3=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\y=6\end{matrix}\right.\)

TH12 : \(\left[{}\begin{matrix}x=-4\\y-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\y=0\end{matrix}\right.\)

Vậy ....