$\begin{array}{l}\underline{\text{Đáp án+Giải thích các bước giải:}}\\4x^2+4x+y^2=24\\↔4x^2+4x+1+y^2=25\\↔(2x+1)^2+y^2=25\\↔(2x+1)^2+y^2=9+16=25+0(do \,\, (2x+1)^2 \,\, lẻ)\\↔\left[ \begin{array}{l}\begin{cases}(2x+1)^2=9\\y^2=16\\\end{cases}\\\begin{cases}(2x+1)^2=25\\y^2=0\\\end{cases}\end{array} \right.\\↔\left[ \begin{array}{l}\begin{cases}\left[ \begin{array}{l}2x+1=3\\2x+1=-3\end{array} \right.\\\left[ \begin{array}{l}y=4\\y=-4\end{array} \right.\\\end{cases}\\\begin{cases}\left[ \begin{array}{l}2x+1=5\\2x+1=-5\end{array} \right.\\y=0\\\end{cases}\end{array} \right.\\↔\left[ \begin{array}{l}\begin{cases}\left[ \begin{array}{l}2x=2\\2x=-4\end{array} \right.\\\left[ \begin{array}{l}y=4\\y=-4\end{array} \right.\\\end{cases}\\\begin{cases}\left[ \begin{array}{l}2x=4\\2x=-6\end{array} \right.\\y=0\\\end{cases}\end{array} \right.\\↔\left[ \begin{array}{l}\begin{cases}\left[ \begin{array}{l}x=1\\x=-2\end{array} \right.\\\left[ \begin{array}{l}y=4\\y=-4\end{array} \right.\\\end{cases}\\\begin{cases}\left[ \begin{array}{l}x=2\\x=-3\end{array} \right.\\y=0\\\end{cases}\end{array} \right.\\Vậy \,\, (x,y)=(1,4),(-2,4),(1,-4),(-2,-4),(2,0),(-3,0)\\\underline{\text{CHÚC BẠN HỌC TỐT}}\\\end{array}$
