\(x^2-kx+k-1=0\)
Theo định lý Viet
\(\Rightarrow\left\{{}\begin{matrix}x_1+x_2=\dfrac{-b}{a}\\x_1x_2=\dfrac{c}{a}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_1+x_2=k\\x_1x_2=k-1\end{matrix}\right.\)
Theo yêu cầu đề bài \(x^2_1x_2+x^2_2x_1=5\)
\(\Leftrightarrow x_1x_2\left(x_1+x_2\right)=5\)
\(\Leftrightarrow\left(k-1\right)k=5\)
\(\Leftrightarrow k^2-k=5\)
\(\Leftrightarrow k^2-k-5=0\)
\(\Delta=b^2-4ac\)
\(\Delta=21\)
\(\Rightarrow\left\{{}\begin{matrix}k_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{1+\sqrt{21}}{2}\\k_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{1-\sqrt{21}}{2}\end{matrix}\right.\)
b)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow x^2_1+x^2_2\ge2\sqrt{x^2_1x^2_2}=2\left|x_1x_2\right|\)
\(\Leftrightarrow x^2_1+x^2_2\ge2\left|k-1\right|\)
Vì \(2\left|k-1\right|\ge0\)
\(\Rightarrow x^2_1+x^2_2\ge0\)
Vậy \(Min_{x^2_1+x^2_2}=0\) khi \(k=1\)