$\begin{array}{l} \text{- Ta có :}\,\,x \div 7 \,\,\text{dư }5 \\\to x=7k+5\,\,(k \in \mathbb{N}) \\\text{- Ta lại có :} \,\,x \div13 \text{ dư }4 \\\to x-4\,\,\vdots\,\,13 \\\to7k+5-4 \,\,\vdots\,\,13 \\\to7k+1 \,\,\vdots\,\,13\\\to7k+1+13\,\,\vdots\,\,13\\\to7k+14\,\,\vdots\,\,13\\\to7(k+2)\,\,\vdots\,\,13 \\\text{mà (7; 13) = 1} \\\to k+2 \,\,\vdots\,\,13\\\to k+2=13m\,\,(m \in \mathbb{N})\\\to k=13m-2 \\\text{- Thay }k=13m-2 \text{ ta có :} \\ x=7k+5 \\\to x=7(13m-2)+5 \\\to x=91m-14+5 \\\to x=91m-9 \\\to x-82=91m-9-82 \\\to x-82=91m-91\,\,\vdots\,\,91\\\to x-82 \,\,\vdots \,\,91 \\\to x-82=91n\,\,(n \in \mathbb{N}) \\\to x=91n+82 \\\to x \div 91 \text{ dư }82 \\\text{- Vậy x chia 91 dư 82}\end{array}$
