钟君申论2016:已知x^2+y^2-4x+y+17/4=0,求y^-x+3xy

来源:百度文库 编辑:高校问答 时间:2024/05/03 11:21:42
已知x^2+y^2-4x+y+17/4=0,求y^-x+3xy
要有步骤

x^2+y^2-4x+y+17/4=0
x^2-4x+4+y^2+y+1/4=0
(x-2)^2+(y+1/2)^2=0
x-2=0,y+1/2=0
x=2,y=-1/2

y^(-x)+3xy=4-3=1

∵x^2+y^2-4x+y+17/4=(x-2)^2+(y+1/2)^2=0
∴x=2,y=-1/2
∴求y^-x+3xy =(-1/2)^-2+3[2*(-1/2)]=....
(自己算一下吧,"3xy" 是指数还是底数,题中分不出!)