已知实数x,y,z满足(绝对值4x-4y+1)+(1\3根号2y+z)+(z的平方-z+1\4)等于0

实数x,y,z满足|4x-4y+1|+(1/3)√（2y+z)+(z^2-z+1/4)=0,
因|4x-4y+1|>=0,(1/3)√（2y+z)>=0,z^2-z+1/4=(z-1/2)^2>=0,
故4x-4y+1=0,
2y+z=0,
z-1/2=0,
解得z=1/2,y=-1/4,x=-1/2.
∴（y+z）x^2=(1/4)(1/4)=1/16.