Question

配方法 已知x2+y2+xy-3y+3=0 求x的y次方

我们初二期末考试的 大概就是这样

Answer 1

原式配方得：x²+xy+y²/4 + 3y²/4-3y+3=0
即得：(x+ y/2)²+ 3(y/2 -1)²=0
则有：x+ y/2=0且y/2 -1=0
解得：y/2=1即y=2，x=-y/2=-1
所以：x的y次幂=(-1)²=1

Answer 2

x^2+y^2+xy-3y+3=0
(x + (1/2)y )^2 + (3/4)y^2 - 3y +3 =0
(x + (1/2)y )^2 + (3/4)( y^2 - 4y +4) =0
(x + (1/2)y )^2 + (3/4)( y -2)^2 =0
x+(1/2)y = 0 and y=2
x= -1 and y=2
x^y
=(-1)^2
=1