用判别式法:先将原式化成一元二次方程,yx^2-ax+2y-b=0,所以 1当y=0,0在值域内 2 判别式=a^2-8y^2+4by大于等于0,即 y^2-b/2y-a^2/8小于等于0,他等价于(y+1)(y-4)小于等于0,即y^2-3y-4小于等于0,所以b/2=3,a^2/8=4,所...