Xét pt hoành độ giao điểm của y = x2 và y = (2m + 1)x - 2 (x \(\ne\) \(\dfrac{1}{2}\))
x2 = (2m + 1)x - 2
\(\Leftrightarrow\) x2 - (2m + 1)x + 2 = 0
\(\Delta\) = [-(2m + 1)]2 - 4.1.2 = 4m2 + 4m + 1 - 8 = 4m2 + 4m - 7
Vì pt có 2 nghiệm x1; x2 \(\Rightarrow\) \(\Delta\) \(\ge\) 0 \(\Leftrightarrow\) m + \(\dfrac{1}{2}\) \(\ge\) \(\pm\)\(\sqrt{2}\) \(\Leftrightarrow\) m \(\ge\) \(\pm\sqrt{2}-\dfrac{1}{2}\)
x1 = \(\dfrac{2m+1+\sqrt{4m^2+4m-7}}{2}\)
x2 = \(\dfrac{2m+1-\sqrt{4m^2+4m-7}}{2}\)
|x1| + |x2| = 4 \(\Leftrightarrow\) \(\dfrac{4m+2}{2}=\pm4\) \(\Leftrightarrow\) 2m + 1 = \(\pm4\) \(\Leftrightarrow\) \(\left[{}\begin{matrix}m=\dfrac{3}{2}\\m=\dfrac{-5}{2}\left(KTM\right)\end{matrix}\right.\)
Vậy ...
x1 = 9x2 \(\Leftrightarrow\) x1 - 9x2 = 0 \(\Leftrightarrow\) x1 + x2 - 10x2 = 0 \(\Leftrightarrow\) 4 - 10x2 = 0
\(\Leftrightarrow\) 10x2 = 4 \(\Leftrightarrow\) x2 = \(\dfrac{2}{5}\) \(\Leftrightarrow\) \(\dfrac{2m+1-\sqrt{4m^2+4m-7}}{2}=\dfrac{2}{5}\)
\(\Leftrightarrow\) 10m + 5 - 5\(\sqrt{4m^2+4m-7}\) = 4
\(\Leftrightarrow\) 1 + 10m = 5\(\sqrt{4m^2+4m-7}\)
\(\Leftrightarrow\) 1 + 20m + 100m2 = 25(4m2 + 4m - 7)
\(\Leftrightarrow\) 1 + 20m + 100m2 - 100m2 - 100m + 175 = 0
\(\Leftrightarrow\) -180m + 176 = 0
\(\Leftrightarrow\) m = \(\dfrac{44}{45}\) (TM)
Chúc bn học tốt! (Phần x1 = 9x2 ko chắc lắm)