\(\Delta=\left(-5\right)^2-4\left(m-1\right)=-4m+4+25=-4m+29\)
Để phương trình có hai nghiệm phân biệt thì -4m+29>0
=>-4m>-29
hay m<29/4
Theo đề, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=5\\2x_1=\sqrt{x_2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1=5-x_2\\x_2=4x_1^2=4\cdot\left(5-x_2\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_1=5-x_2\\4x_2^2-40x_2+100-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1=5-x_2\\\left[{}\begin{matrix}x=\dfrac{11}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left(x_1;x_2\right)\in\left\{\left(-\dfrac{1}{2};\dfrac{11}{2}\right);\left(\dfrac{1}{2};\dfrac{9}{2}\right)\right\}\)
Theo đề, ta có: \(x_1x_2=m-1\)
\(\Leftrightarrow\left[{}\begin{matrix}m-1=-\dfrac{11}{4}\\m-1=\dfrac{9}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=-\dfrac{7}{4}\left(nhận\right)\\m=\dfrac{13}{4}\left(nhận\right)\end{matrix}\right.\)
