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y=(x²-2x+3)/(2x-3)函数的值域
人气:139 ℃ 时间:2020-01-26 09:04:04
解答
x^2-2x+3=2yx-3y
x^2-(2+2y)x+(3+3y)=0
判别式:
4(y+1)^2-12(y+1)≥0
4(y+1)(y-2)≥0
y≥2或y≤-1函数y=(1+√x)/(1-√x)的值域是1+√x=y-y√x
(1+y)√x=(y-1)
√x=(y-1)/(y+1)
因为√x≥0
(y-1)/(y+1)≥0
(y-1)(y+1)≥0(y≠-1)
y≥1,或y<-1
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