10: \(y=x^3+2\left(m-1\right)x^2+\left(m^2-4m+1\right)x-2\left(m^2+1\right)\)
=>y'=\(3x^2+2\left(m-1\right)\cdot2x+\left(m^2-4m+1\right)\)
=>y'=\(3x^2+\left(4m-4\right)x+m^2-4m+1\)
Để hàm số có hai cực trị thì phương trình y'=0 có hai nghiệm phân biệt
=>\(\left(4m-4\right)^2-4\cdot3\cdot\left(m^2-4m+1\right)>0\)
=>\(16m^2-32m+16-12m^2+48m-12>0\)
=>\(4m^2+16m+4>0\)
=>\(m^2+4m+1>0\)
=>\(m^2+4m+4>3\)
=>\(\left(m+2\right)^2>3\)
=>\(\left[\begin{array}{l}m+2>\sqrt3\\ m+2<-\sqrt3\end{array}\right.\Rightarrow\left[\begin{array}{l}m>\sqrt3-2\\ m<-\sqrt3-2\end{array}\right.\)
Theo Vi-et, ta có:
\(\begin{cases}x_1+x_2=-\frac{b}{a}=\frac{-4m+4}{3}\\ x_1x_2=\frac{c}{a}=\frac{m^2-4m+1}{3}\end{cases}\)
\(\frac{1}{x_1}+\frac{1}{x_2}=\frac12\left(x_1+x_2\right)\)
=>\(\frac{x_1+x_2}{x_1x_2}-\frac12\left(x_1+x_2\right)=0\)
=>\(\left(x_1+x_2\right)\left(\frac{1}{x_1x_2}-\frac12\right)=0\)
TH1: x1+x2=0
=>-4m+4=0
=>-4m=-4
=>m=1(nhận)
TH2: \(\frac{1}{x_1x_2}-\frac12=0\)
=>\(x_1x_2=2\)
=>\(\frac{m^2-4m+1}{3}=2\)
=>\(m^2-4m+1=6\)
=>\(m^2-4m-5=0\)
=>(m-5)(m+1)=0
=>m=5(nhận) hoặc m=-1(loại)
