a/ Bạn tự giải
b/ Pt hoành độ giao điểm: \(x^2-4x+m-1=0\)
\(\Delta'=4-m+1=5-m\ge0\Rightarrow m\le5\)
\(\sqrt{y_1}\sqrt{y_2}=5\Leftrightarrow\sqrt{x_1^2}\sqrt{x_2^2}=5\)
\(\Leftrightarrow\left|x_1\right|\left|x_2\right|=5\Leftrightarrow\left|x_1x_2\right|=5\)
\(\Leftrightarrow\left|m-1\right|=5\Rightarrow\left[{}\begin{matrix}m=6\left(l\right)\\m=-4\end{matrix}\right.\)