\(\Delta'=n^2-6n+9-3n^2+8n-5=-2n^2+2n+4\)
Để phương trình có hai nghiệm phân biệt thì \(\Delta'>0\Leftrightarrow-2n^2+2n+4>0\)
\(\Rightarrow\left\{{}\begin{matrix}n>-1\\n>2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}n>-1\\n\ne2\end{matrix}\right.\)
Theo đề:
\(x_1^2+x_2+2x_2^2-x_1=3x_1x_2\)
\(\Leftrightarrow\left(x_1^2-x_1x_2\right)-\left(2x_1x_2-2x_2^2\right)=x_1-x_2\)
\(\Leftrightarrow\left(x_1-x_2\right)x_1-2x_2\left(x_1-x_2\right)-\left(x_1-x_2\right)=0\)
\(\Leftrightarrow\left(x_1-x_2\right)\left(x_1-2x_2-1\right)=0\Leftrightarrow x_1-2x_2=1\)
Kết hợp vi ét ta có: \(\left\{{}\begin{matrix}x_1-2x_2=1\left(1\right)\\x_1+x_2=2n-6\left(2\right)\\x_1x_2=3n^2-8n+5\left(3\right)\end{matrix}\right.\)
Từ (1) và (2) có \(x_1=1+2x_2\) \(\Rightarrow1+2x_2+x_2=2n-6\Rightarrow x_2=\dfrac{2n-6-1}{3}=\dfrac{2n-7}{3}\)
\(\Rightarrow x_1-\dfrac{2.\left(2n-7\right)}{3}=1\Rightarrow\dfrac{3}{3}+\dfrac{4n-14}{3}=\dfrac{4n-11}{3}\)
Từ (3) có: \(\left(\dfrac{2n-7}{3}\right)\left(\dfrac{4n-11}{3}\right)=3n^2-8n+5\)
\(\Leftrightarrow\dfrac{8n^2-50n+77}{9}-\dfrac{27n^2-72n+45}{9}=0\)
\(\Leftrightarrow-19n^2+22n+32=0\)
\(\Rightarrow\left\{{}\begin{matrix}n=-\dfrac{16}{19}\left(nhận\right)\\n=2\left(loại\right)\end{matrix}\right.\)
Vậy giá trị n cần tìm là \(\dfrac{-16}{19}\)
