已知集合A={（X,Y）|Y=-X^2+mx-1},B={(x,y)|x+y=3,0

已知集合A={（X,Y）|Y=-X^2+mx-1},B={(x,y)|x+y=3,0

相当于线段：x+y=3,0=

依题意有，3-x=-x^2+mx-1,在0<=x<=3上有唯一实数解。由，3-x=-x^2+mx-1得x^2-(m+1)x+4=0
设f(x）=x^2-(m+1)x+4,所以有，f(0)*f(3)<=0,即4×[3^2-(m+1)×3+4]<=0,解得m>=10/3.