若k为4,且根号下(4k+y方)-根号下(5k-y方)等于根号下7,求根号下(4k+y方)+根号下(5k-y方)的值.
来源:百度知道 编辑:UC知道 时间:2024/07/08 02:27:45
根号下(4k+y方)-根号下(5k-y方)等于根号下7
二边同时平方得:
4k+y^2-2根号(4k+y^2)*根号(5k-y^2)+5k-y^2=7
所以:2根号(4k+y^2)*根号(5k-y^2)=9k-7=36-7=29
所求原式的平方=4k+y^2+2根号(4k+y^2)*根号(5k-y^2)+5k-y^2=9k+29=9*4+29=65
那么所求原式=根号65
k=4
根号(4k+y^2)-根号(5k-y^2)=根号(14+y^2)-根号(20-y^2)=根号7
两边平方,则
(14+y^2)+(20-y^2)-2根号(14+y^2)(20-y^2)=7
34-2根号(14+y^2)(20-y^2)=7
根号(14+y^2)(20-y^2)=27/2
(14+y^2)+(20-y^2)+2根号(14+y^2)(20-y^2)=34+27=61
[根号(14+y^2)+根号(20-y^2)]^2=61
根号下(4k+y方)+根号下(5k-y方)=根号61
(4k+y^2)^0.5-(5k-y^2)^0.5=7^0.5
两边平方,得
4k+y^2+5k-y^2-2[(4k+y^2)(5k-y^2)]^0.5=7
k=4,得,2[(4k+y^2)(5k-y^2)]^0.5=29
(4k+y^2)^0.5-(5k-y^2)^0.5
两边平方,得
4k+y^2+5k-y^2-2[(4k+y^2)(5k-y^2)]^0.5=7
k=4,得,2[(4k+y^2)(5k-y^2)]^0.5=29
4k+y^2+5k-y^2+2[(4k+y^2)(5k-y^2)]^0.5
k=4,得,9k+2[(4k+y^2)(5k-y^2)]^0.5=36+29=65
所以(4k+y^2)^0.5-(5k-y^2)^0.5=65^0.5或-65^0.5
因为(4k+y^2)^0.5+(5k-y^2)^0.5>(4k+y^2)^0.5-(5k-y^2)^0.