a: \(x^4+3x^2-4=0\)
=>\(x^4+4x^2-x^2-4=0\)
=>\(\left(x^2+4\right)\left(x^2-1\right)=0\)
=>\(x^2-1=0\)
=>\(x^2=1\)
=>\(\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
b: \(2x^2-6x+3m+1=0\)
\(\text{Δ }=\left(-6\right)^2-4\cdot2\cdot\left(3m+1\right)\)
\(=36-8\left(3m+1\right)\)
\(=36-24m-8=-24m+28\)
Để phương trình có hai nghiệm thì Δ>=0
=>-24m+28>=0
=>-24m>=-28
=>\(m< =\dfrac{7}{6}\)
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{-\left(-6\right)}{2}=\dfrac{6}{2}=3\\x_1x_2=\dfrac{c}{a}=\dfrac{3m+1}{2}\end{matrix}\right.\)
\(x_1^3+x_2^3=9\)
=>\(\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)=9\)
=>\(3^3-3\cdot\dfrac{3m+1}{2}\cdot3=9\)
=>\(9\cdot\dfrac{3m+1}{2}=27-9=18\)
=>\(3m+1=\dfrac{18}{4,5}=4\)
=>m=1(nhận)
