Bài 2
Hoành độ giao điểm tm pt
\(\dfrac{1}{2}x^2=mx-\dfrac{1}{2}m^2+m+1\Leftrightarrow x^2-2mx+m^2-m-1=0\)
Thay m = 1 ta được
\(x^2-2x-1=0\Leftrightarrow\left(x-1\right)^2-2=0\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}+1\\x=-\sqrt{2}+1\end{matrix}\right.\)
Ta có \(\Delta'=m^2-\left(m^2-m-1\right)=m+1\)
Để pt có 2 nghiệm pb m > -1
Theo Vi et \(\left\{{}\begin{matrix}x_1+x_2=2m\\x_1x_2=m^2-m-1\end{matrix}\right.\)
Ta có |x1-x2|=2 <=> \(\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}=2\)
\(\Leftrightarrow\sqrt{4m^2-4m^2+4m+4}=2\Leftrightarrow2\sqrt{m+1}=2\Leftrightarrow m=0\left(tm\right)\)
Bài 1
\(\left\{{}\begin{matrix}3x+3+2x+4y=4\\4x+4-x-2y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x+4y=1\\3x-2y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x+4y=1\\6x-4y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
