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问题：【y=1/x平方+根号x的值域怎么求?】
答案：↓↓↓ 程显蒙的回答： 网友采纳 y=1/(x^2)+x^(1/2) 由于存在根号,所以x大于零 y=1/(x^2)+(1/4)*x^(1/2))+(1/4)*x^(1/2))+(1/4)*x^(1/2))+(1/4)*x^(1/2) >=5*五次根号下(1/(x^2)*(1/4)*x^(1/2))*(1/4)*x^(1/2))*(1/4)*x^(1/2))*(1/4)*x^(1/2)) =5*（2的负五分之八次方） 当1/(x^2)＝(1/4)*x^(1/2)时取等号,即x＝2的负五分之四次方时有最小值5*（2的负五分之八次方）,最大是正无穷大 | 
