
y=2x-1/x^2+2x+2和y=x-2/x^2-3x+2的值域
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2014-07-26 · 知道合伙人教育行家
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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,+∞)
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,+∞)
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