已知实数x,y,z满足|4x-4y+1|+(1/3)√(2y+z)+(z²-z+1/4)=0。求(y+z)x²的值。
解:将原式化为|4x-4y+1|+(1/3)√(2y+z)+[z-(1/2)]²=0,可知:
4x-4y+1=0...........(1)
2y+z=0................(2)
z-1/2=0...............(3)
由(3)得z=1/2;代入(2)式得y=-(1/2)z=-1/4;代入(1)式得x=(1/4)(4y-1)=-1/2;
∴(y+z)x²=(-1/4+1/2)(-1/2)²=1/16
|4x-4y+1|+1/3√2y+z+(z-1/2)^2=0
由几个非负数的和为零,要求每一项为零
故4x-4y+1=0
2y+z =0
z-1/2=0
x=-1/2
y=-1/4
z=1/2
所求为(y+z)*x^2=1/4*1/4=1/16
题目有问题吧?