y=2x-1/x^2+2x+2和y=x-2/x^2-3x+2的值域
1个回答

y=(2x-1)/(x^2+2x+2)

分母x^2+2x+2=(x+1)^2+1>0

y(x^2+2x+2)= 2x-1

yx^2+2yx+2y= 2x-1

yx^2+2(y-1)x+(2y+1)= 0

判别式△=4(y-1)^2-4y(2y+1)=4y^2-8y+4-8y^2-4y=-4y^2-12y+4≥0

y^2+3y-1≤0

(-3-√13)/2≤y≤(-3+√13)/2

值域【(-3-√13)/2,(-3+√13)/2】

y=(x-2)/(x^2-3x+2)

= (x-2)/[(x-1)(x-2)]

= 1/(x-1)

1/(x-1)≠0

值域(-∞,0),(0,+∞)