y=(2x^2+8x+6)/(x^2+4x+5)求值域
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y=(2x^2+8x+6)/(x^2+4x+5)求值域y=(2x^2+8x+6)/(x^2+4x+5)求值域y=(2x^2+8x+6)/(x^2+4x+5)求值域y=2-4/(x^2+4x+5);而4
y=(2x^2+8x+6)/(x^2+4x+5)求值域
y=(2x^2+8x+6)/(x^2+4x+5)求值域
y=(2x^2+8x+6)/(x^2+4x+5)求值域
y=2- 4/(x^2+4x+5);而4/(x^2+4x+5)是大于0而小于等于4
故y=(2x^2+8x+6)/(x^2+4x+5)的值域【-2,2)
不懂再问
(y-2)x²+(4y-8)x+5y-6=0
若y=2,则 上式左边=4≠0,所以y≠2
y≠2,将(y-2)t²+(4y-8)t+5y-6=0看成t的二次方程因其在实数范围内有解x
则其判别式Δ=(4y-8)²-4(y-2)(5y-6)≥0
(4y-8)²-4(y-2)(5y-6)=16y²-64y+64-4(5...
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(y-2)x²+(4y-8)x+5y-6=0
若y=2,则 上式左边=4≠0,所以y≠2
y≠2,将(y-2)t²+(4y-8)t+5y-6=0看成t的二次方程因其在实数范围内有解x
则其判别式Δ=(4y-8)²-4(y-2)(5y-6)≥0
(4y-8)²-4(y-2)(5y-6)=16y²-64y+64-4(5y²-16y+12)=-4y²+16≥0
解得 -2≤y≤2
因条件有y≠2,所以有-2≤y<2
即y的值域为[-2,2)
收起
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