Question

Cho phương trình: \(x^2\) - mx + m - 2 = 0 (m là tham số). Tìm m để phương trình vó 2 nghiệm \(x_1\), \(x_2\) thỏa mãn: \(\frac{x_1^2-2}{x_1-1}\).\(\frac{x_2^2-2}{x_2-1}\) = 4

Answer 1

\(\Delta=m^2-4\left(m-2\right)=\left(m-2\right)^2+4>0;\forall m\)

\(\left\{{}\begin{matrix}x_1+x_2=m\\x_1x_2=m-2\end{matrix}\right.\)

\(\frac{\left(x_1^2-2\right)\left(x_2^2-2\right)}{\left(x_1-1\right)\left(x_2-1\right)}=4\Leftrightarrow\frac{\left(x_1x_2\right)^2-2\left(x_1+x_2\right)^2+4}{x_1x_2-\left(x_1+x_2\right)+1}=4\)

\(\Leftrightarrow\frac{\left(x_1x_2\right)^2-2\left(x_1+x_2\right)^2+4x_1x_2+4}{x_1x_2-\left(x_1+x_2\right)+1}=4\)

\(\Leftrightarrow\left(m-2\right)^2-2m^2+4\left(m-2\right)+4=4\left(m-2\right)-4m+4\)

\(\Leftrightarrow4-m^2=0\Rightarrow m=\pm2\)