Theo Vi-et, ta có:
\(x_1+x_2=-\dfrac{b}{a}=4;x_1x_2=\dfrac{c}{a}=3\)
\(5m^2-\left(x_1^3\cdot x_2+x_1\cdot x_2^3\right)\cdot m+25=0\)
=>\(5m^2-\left[x_1x_2\left(x_1^2+x_2^2\right)\right]\cdot m+25=0\)
=>\(5m^2-\left[3\cdot\left[\left(x_1+x_2\right)^2-2x_1x_2\right]\right]\cdot m+25=0\)
=>\(5m^2-\left[3\cdot\left(4^2-2\cdot3\right)\right]\cdot m+25=0\)
=>\(5m^2-30m+25=0\)
=>\(m^2-6m+5=0\)
=>(m-1)(m-5)=0
=>\(\left[{}\begin{matrix}m=1\\m=5\end{matrix}\right.\)
Theo Viet: \(\left\{{}\begin{matrix}x_1+x_2=4\\x_1x_2=3\end{matrix}\right.\)
\(x_1^3x_2+x_1x_2^3=\left(x_1x_2\right)\left(x_1^2+x_2^2\right)=x_1x_2\left[\left(x_1+x_2\right)^2-2x_1x_2\right]\)
\(=3.\left(4^3-2.3\right)=30\)
Pt trở thành:
\(5m^2-30m+25=0\Rightarrow\left[{}\begin{matrix}m=1\\m=5\end{matrix}\right.\)
