PT hoành độ giao điểm: \(2x^2=-2mx+m+1\)
\(\Leftrightarrow2x^2+2mx-\left(m+1\right)=0\)
Vì (P) cắt (d) tại 2 điểm phân biệt nên \(\Delta'=m^2+2\left(m+1\right)>0\)
\(\Leftrightarrow\left(m+1\right)^2>0\left(\text{đúng với mọi }m\ne-1\right)\)
Áp dụng Viét: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{2m}{2}=-m\\x_1x_2=\dfrac{-\left(m+1\right)}{2}\end{matrix}\right.\)
Ta có \(\dfrac{1}{\left(2x_1-1\right)^2}+\dfrac{1}{\left(2x_2-1\right)^2}=2\)
\(\Leftrightarrow\dfrac{4x_2^2-4x_2+1+4x_1^2-4x_1+1}{\left[\left(2x_1-1\right)\left(2x_2-1\right)\right]^2}=2\\ \Leftrightarrow4\left[\left(x_1+x_2\right)^2-2x_1x_2\right]-4\left(x_1+x_2\right)+2=2\left[4x_1x_2-2\left(x_1+x_2\right)+1\right]^2\\ \Leftrightarrow4\left(m^2+m+1\right)+4m=2\left(-2m-2+2m+1\right)^2\\ \Leftrightarrow4m^2+4m+4+4m=2\\ \Leftrightarrow2m^2+4m+1=0\\ \Leftrightarrow\left[{}\begin{matrix}m=\dfrac{-2+\sqrt{2}}{2}\left(tm\right)\\m=\dfrac{-2-\sqrt{2}}{2}\left(tm\right)\end{matrix}\right.\)
