\(a,\left(x-1\right)\left(x^2+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x^2+1=0\end{matrix}\right..Mà:x^2+1>0nên:x-1=0\Leftrightarrow x=1\left(tm\right).Vậy:x=1\)
\(b,\left|x+5\right|\ge0\Rightarrow\left|x+5\right|+3\ge3\left(\text{ không tìm được x thỏa mãn}\right)\)
\(c,y^2-9=0\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right..Vậy:x\in\left\{-3;3\right\}\)
\(d,\frac{2x+1}{-3}=0\Leftrightarrow2x+1=0\Leftrightarrow2x=-1\Leftrightarrow x=-\frac{1}{2}.Vậy:x=-\frac{1}{2}\)