\(a+b=-10\\↔a=-b-10\)
Thay \(a=-b-10\) vào \(a.b=24\)
\(→(-b-10).b=24\\↔-b^2-10b=24\\↔-b^2-10b-24=0\\↔b^2+10b+24=0\\↔b^2+6b+4b+24=0\\↔(b^2+6b)+(4b+24)=0\\↔b(b+6)+4(b+6)=0\\↔(b+4)(b+6)=0\\↔\left[\begin{array}{1}b+4=0\\b+6=0\end{array}\right.\\↔\left[\begin{array}{1}b=-4\\b=-6\end{array}\right.\)
Với \(b=-4→a=4-10=-6\)
Với \(b=-6→a=6-10=-4\)
mà \(a,b\in\Bbb Z\)
\(→(a,b)=\{(-6;-4);(-4;-6)\}\)
Vậy \( (a,b)=\{(-6;-4);(-4;-6)\}\)
