Question

已知(x-2/3)^2+丨y+1丨=0试求二分之一x-2(x-1/3y^2)+(1/3y^2-3/2x)的值

Answer 1

已知(x-2/3)^2+丨y+1丨=0
则x-2/3=0
y+1=0
解得x=2/3
y=-1
所以二分之一x-2(x-1/3y^2)+(1/3y^2-3/2x)
=(1/2)*(2/3)-2[2/3-(1/3)(-1)²]+(1/3)*(-1)²-(3/2)*(2/3)
=1/3-2*(1/3)+1/3-1
=-1

Answer 2

解：因为(x-2/3)^2+丨y+1丨=0
且(x-2/3)^2≥0，丨y+1丨≥0
所以x=2/3，y=-1
1/2x-2(x-1/3y^2)+(1/3y^2-3/2x)
=1/2x-2x+2/3y^2+1/3y^2-3/2x
=-3x+y^2
当x=2/3，y=-1时
原式=-3×2/3+（-1）²=-2+1=-1