\(a.a-b=2\Rightarrow a=b+2\)
Ta có : \(ab=15\Leftrightarrow\left(b+2\right)b=15\Leftrightarrow b^2+2b-15=0\Leftrightarrow b^2-3b+5b-15=0\Leftrightarrow\left(b+5\right)\left(b-3\right)=0\Leftrightarrow\left[{}\begin{matrix}b=-5\\b=3\end{matrix}\right.\)
Khi đó : \(a=\left[{}\begin{matrix}-5+2=-3\\3+2=5\end{matrix}\right.\)
\(b.a+b=-8\Leftrightarrow a=-b-8\)
Ta có : \(ab=15\Leftrightarrow-\left(b+8\right)b=15\Leftrightarrow-b^2-8b-15=0\Leftrightarrow-b^2-3b-5b-15=0\Leftrightarrow-\left(b+3\right)\left(b+5\right)=0\Leftrightarrow\left[{}\begin{matrix}b=-3\\b=-5\end{matrix}\right.\)
Khi đó : \(a=\left[{}\begin{matrix}-5-8=-13\\-3-8=-11\end{matrix}\right.\)
Thử lại thấy không thỏa mãn .
Vậy , ............
