a)A= x2-10=-3x
⇔x2+3x-10=0
⇔x2+5x-2x-10=0
⇔(x2+5x)-(2x+10)=0
⇔x(x+5)-2(x+5)=0
⇔(x+5)(x-2)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x+5=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
vậy pt A có tập no S={-5;2}
b)
B=\(\dfrac{A}{x^2+10}=\dfrac{x^2-10}{x^2+10}=\dfrac{x^2+10-20}{x^2+10}=1-\dfrac{20}{x^2+10}\)
Do \(x^2\ge0\forall x\)
=>\(x^2+10\ge10\)
=>\(\dfrac{20}{x^2+10}\le2\)
=>\(-\dfrac{20}{x^2+10}\ge-2\)
=>\(1-\dfrac{20}{x^2+10}\ge-1\)
=> B\(\ge-1\)
=> GTNN B=-1