\(x^2-2\left(n-1\right)x-n-5=0\) (1)
\(\Delta=\left[-2\left(n-1\right)\right]^2-4\cdot1\cdot\left(-n-5\right)=4\left(n-1\right)^2+4\left(n+5\right)\)
\(=4\left(n^2-2n+1\right)+4n+20=4n^2-4n+1+23=\left(2n-1\right)^2+23>0\forall x\)
Pt luôn có 2 nghiệm phân biệt
Theo vi-ét: \(\left\{{}\begin{matrix}x_1+x_2=2\left(n-1\right)\\x_1x_2=-n-5\end{matrix}\right.\)
Ta có:
\(x_1^2+x_2^2=14\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=14\)
\(\Rightarrow\left[2\left(n-1\right)\right]^2-2\cdot\left(-n-5\right)=14\)
\(\Leftrightarrow4\left(n-1\right)^2+2\left(n+5\right)=14\)
\(\Leftrightarrow4n^2-8n+4+2n+10=14\)
\(\Leftrightarrow4n^2-6n=0\)
\(\Leftrightarrow2n\left(2n-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}n=0\\2n=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}n=0\\n=\dfrac{3}{2}\end{matrix}\right.\)
Vậy: ...