设二次函数与直线的两个交点为A（x1,y1),B(x2,y2)
则y1=-x1+1 y2=-x2+1
AB^2=(X1-X2)^2+(Y1-Y2)^2
Y1-Y2=-X1+1+X2-1=X2-X1
(Y1-Y2)^2=(X2-X1)^2=(X1-X2)^2
所以AB^2=(X1-X2)^2+(Y1-Y2)^2=2(X1-X2)^2=2√2^2=8
(X1-X2)^2=4,变形后得：
(X1+X2)^2-4X1X2=4
联立y=(m+1)x^2/2-(3m+1)x/2+m和y=-x+1
(m+1)x^2/2-(3m+1)x/2+m=-x+1
(m+1)x^2/2-（3m-1）x/2+m-1=0
x1+x2=（3m-1）/(m+1)
x1x2=2(m-1)/(m+1)
代入上式得
（3m-1）^2/(m+1)^2-8(m-1)/(m+1)=4
m=-5 or 1/3
因为delta>0
所以 (（3m-1）/2)^2-4*(m+1)/2 * (m-1)>0
M^2-6M+9＞0
M≠3
所以m=-5 or 1/3