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求下列函数的值域 (1)Y=(5X-1)/(4X+2) (2)Y=(X^2-4X+3)/(2x^2-x-1) (3)Y=(2X^2+4X-7)/(X^2+2X+3)
(4)Y=2X-√(X-1)
(5)Y=4-√(3+2X-X^2)
人气:487 ℃ 时间:2019-10-11 14:36:07
解答
(1)Y=(5X-1)/(4X+2)=5/4 * (x+1/2-1/2-1/5)/(x-1/2) 【x≠1/2】=5/4[(x+1/2)-7/10]/(x-1/2)=5/4-(7/8)/(x-1/2)≠5/4值域:(-∞,5/4),(5/4,+∞)(2)Y=(X^2-4X+3)/(2x^2-x-1) =[(x-1)(x-3)]/[(2x+1)(x-1)] ...
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