a) (x+2)(y-3)=0

\(\orbr{\begin{cases}x+2=0\\y-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\y=3\end{cases}}}\)

a/  (x + 2) . ( y - 3 ) = 0 

Nên: x + 2 = 0 => x = 0 - 2 = - 2 

Hoặc y - 3 = 0 => y = 0+ 3 = 3

Vậy x = -2 hoặc y = 3

b/ (x + 4) . ( y - 2 ) = 2

Nên:

(-) x + 4 = 1 => ..........

   y - 2  = 2 => ....

(-) x + 4 = 2 => ....

    y - 2 =  1 => ....

(-) x + 4 = -1 =>....

    y - 2 = -2 => ....

(-) x + 4 = -2

    y -  2 = -1 => .......  

* Các trường hợp kia tương tự

x+y=-10

=>x=-10-y

xy=24

\(\Leftrightarrow y\left(-10-y\right)=24\)

\(\Leftrightarrow y^2+10y+24=0\)

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

3-x và 7-y € Ư(5)

Mà Ư(5)= {1;5;-1;-5}

Ta có bảng sau:

4-x và y+3 € Ư(-3)

Mà Ư(-3)={1;3;-1;-3}

Ta có bảng sau:

\(\frac{3+x}{7+y}=\frac{3}{7};x+y=20\)

\(\Leftrightarrow21+7x=21+3y\Leftrightarrow7x=3y\Leftrightarrow\frac{x}{3}=\frac{y}{7}\)

Áp dụng t/c dãy tỉ số ''='' nhau ta có 

\(\frac{x}{3}=\frac{y}{7}=\frac{x+y}{3+7}=\frac{2}{10}=\frac{1}{5}\)

\(\Leftrightarrow\frac{x}{3}=\frac{1}{5}\Leftrightarrow5x=3\Leftrightarrow x=\frac{3}{5}\)

\(\Leftrightarrow\frac{y}{7}=\frac{1}{5}\Leftrightarrow5y=7\Leftrightarrow y=\frac{7}{5}\)

1)\(\left(x+1\right).\left(y-2\right)=0\)                                       \(\left(x,y\inℤ\right)\)

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

2)\(\left(x-5\right).\left(y-7\right)=1\)

3)\(\left(x+4\right).\left(y-2\right)=2\)

4)\(\left(x-4\right).\left(y+3\right)=-3\)

5)\(\left(x+3\right).\left(y-6\right)=-4\)

6)\(\left(x-8\right).\left(y+7\right)=5\)

7)\(\left(x+7\right).\left(y-3\right)=-6\)

8)\(\left(x-6\right).\left(y+2\right)=7\)

ok :)