Phương trình hoành độ giao:
$x^2-x+m=0$
Hai giao điểm phân biệt khi $\Delta>0$
$\Delta=1-4m=-4m+1>0$
$\Leftrightarrow m<\dfrac{1}{4}$
Theo Viet: $x_1+x_2=1; x_1x_2=m$
$x_1+x_2=1$
Ta có:
$|x_2-x_1|=\sqrt{(x_2-x_1)^2}=\sqrt{x_2^2+x_1^2-2x_1x_2}=\sqrt{(x_2+x_1)^2-4x_1x_2}$
$\Rightarrow (x_2-x_1)^4=[(x_2+x_1)^2-4x_1x_2]^2=(1-4m)^2=(4m-1)^2$
$|x_2-x_1|=\sqrt{1-4m}$
$\Rightarrow (x_2+x_1)|x_2-x_1|=\sqrt{1-4m}$
$\Leftrightarrow [(x_2+x_1)(x_2-x_1)]^2=1-4m$
$\Leftrightarrow (x_2^2-x_1^2)^2=1-4m$
$\Leftrightarrow (y_2-y_1)^4=(1-4m)^2=(4m-1)^2$
Do đó ta có:
$(4m-1)^2+(4m-1)^2=18$
$\Leftrightarrow (4m-1)^2=9$
$\Leftrightarrow |4m-1|=3$
$\Leftrightarrow m=1$ (loại) hoặc $m=\dfrac{-1}{2}$ (TM)
Vậy $m=\dfrac{-1}{2}$
