bài 3:
a: Theo đề, ta có hệ:
\(\left\{{}\begin{matrix}a-4+c=3\\16a+16+c=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{19}{15}\\c=\dfrac{124}{15}\end{matrix}\right.\)
b: Theo đề, ta có:
\(\left\{{}\begin{matrix}\dfrac{4}{2a}=2\\a+4+c=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\c+4=4\end{matrix}\right.\Leftrightarrow a=1;c=0\)
c: Theo đề, ta có:
\(\left\{{}\begin{matrix}\dfrac{4}{2a}=4\\-\dfrac{b^2-4ac}{4a}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\\dfrac{16-4\cdot1\cdot c}{4}=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=1\\16-4c=-8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\c=6\end{matrix}\right.\)