Cho pt :x^2-(m+1)x +m^2 -2m +2 =0. Tìm m để pt có 2 nghiệm sao cho A=x1^2+x2^2 min,max 19/08/2021 Bởi Athena Cho pt :x^2-(m+1)x +m^2 -2m +2 =0. Tìm m để pt có 2 nghiệm sao cho A=x1^2+x2^2 min,max
\(\Delta=\left(m+1\right)^2-4\left(m^2-2m+2\right)\ge0\) \(\Leftrightarrow-3m^2+10m-7\ge0\Leftrightarrow1\le m\le\frac{7}{3}\) Khi đó pt có 2 nghiệm thỏa: \(\left\{{}\begin{matrix}x_1+x_2=m+1\\x_1x_2=m^2-2m+2\end{matrix}\right.\) \(A=x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2\) \(A=\left(m+1\right)^2-2\left(m^2-2m+2\right)\) \(A=-m^2+6m-3\) \(A=-m^2+6m-5+2=\left(m-1\right)\left(5-m\right)+2\ge2\) ; \(\forall m\in\left[1;\frac{7}{3}\right]\) \(A_{min}=2\) khi \(m=1\) \(A=-m^2+6m-\frac{77}{9}+\frac{50}{9}=\left(m-\frac{7}{3}\right)\left(\frac{11}{3}-m\right)+\frac{50}{9}\le\frac{50}{9}\) \(A_{max}=\frac{50}{9}\) khi \(m=\frac{7}{3}\) Bình luận
\(\Delta=\left(m+1\right)^2-4\left(m^2-2m+2\right)\ge0\)
\(\Leftrightarrow-3m^2+10m-7\ge0\Leftrightarrow1\le m\le\frac{7}{3}\)
Khi đó pt có 2 nghiệm thỏa: \(\left\{{}\begin{matrix}x_1+x_2=m+1\\x_1x_2=m^2-2m+2\end{matrix}\right.\)
\(A=x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2\)
\(A=\left(m+1\right)^2-2\left(m^2-2m+2\right)\)
\(A=-m^2+6m-3\)
\(A=-m^2+6m-5+2=\left(m-1\right)\left(5-m\right)+2\ge2\) ; \(\forall m\in\left[1;\frac{7}{3}\right]\)
\(A_{min}=2\) khi \(m=1\)
\(A=-m^2+6m-\frac{77}{9}+\frac{50}{9}=\left(m-\frac{7}{3}\right)\left(\frac{11}{3}-m\right)+\frac{50}{9}\le\frac{50}{9}\)
\(A_{max}=\frac{50}{9}\) khi \(m=\frac{7}{3}\)