\(pt\) \(hoành\) \(độ\) \(giao\) \(điểm:-x^2=2\left(2m+1\right)x-1\)
\(\Leftrightarrow-x^2-2\left(2m+1\right)x+1=0\)
\(\Leftrightarrow x^2+2\left(2m+1\right)x-1=0\)
\(\Delta'>0\Leftrightarrow\left(2m+1\right)^2+4>0\left(đúng\forall m\right)\)
\(vi-ét\Rightarrow\left\{{}\begin{matrix}x1+x2=-2\left(2m+1\right)\\x1x2=-1\end{matrix}\right.\)
\(\Rightarrow A\left(x1;x1^2+2\left(2m+1\right)x1-1\right)\)
\(B\left(x2;x2^2+2\left(2m+1\right)x2-1\right)\)
\(\Rightarrow y1+y2+2\left(x1+x2\right)-16m^2=10\)
\(\Leftrightarrow x1^2+2\left(2m+1\right)x1-1+x2^2+2\left(2m+1\right)x2-1+2\left(x1+x2\right)-16m^2-10=0\)
\(\Leftrightarrow\left(x1+x2\right)^2-2x1x2+2\left(2m+1\right)\left(x1+x2\right)+2\left(x1+x2\right)-16m^2-12=0\)
\(\Leftrightarrow4\left(2m+1\right)^2-4\left(2m+1\right)^2+10-16m^2=0\Leftrightarrow16m^2=10\Leftrightarrow m=\pm\sqrt{\dfrac{10}{16}}\)
(kiểm tra lại biến đổi với tính toán giùm nhé)
