a)
vì 5(x+2)(x+3)=1
=> 5(x+2)(x+3)=50
=> (x+2)(x+3)=0
\(\Rightarrow\left[{}\begin{matrix}x+2=0\Rightarrow x=-2\\x+3=0\Rightarrow x=-3\end{matrix}\right.\)
vậy \(x\in\left\{-2;-3\right\}\)
b)
ta có \(\left\{{}\begin{matrix}\left(x-4\right)^{100}\ge0\forall x\\\left|3-y\right|\ge0\forall y\end{matrix}\right.\)
mà (x-4)100+|3-y|=0
\(\Rightarrow\left\{{}\begin{matrix}x-4=0\Rightarrow x=4\\3-y=0\Rightarrow y=3\end{matrix}\right.\)
vậy (x;y) = (4;3)