∮5x/(1-x05)∧1/2
来源:蜘蛛抓取(WebSpider)
时间:2017-12-18 02:24
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答:y=(^2+8x+5)/(x^2+1) 定义域为实数范围R整理得:yx^2+y=^2+8x+5(5-y)x^2+8x+5-y=0方程恒有解判别式>=0所以:8^2-4*(5-y)*(5-y)>=0整理得:(y-5)^2<=16-4<=y-5<=41<=y<=9所以:值域为[149]...
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答:y=(^2+8x+5)/(x^2+1) 定义域为实数范围R整理得:yx^2+y=^2+8x+5(5-y)x^2+8x+5-y=0方程恒有解判别式>=0所以:8^2-4*(5-y)*(5-y)>=0整理得:(y-5)^2<=16-4<=y-5<=41<=y<=9所以:值域为[149]