Question

若绝对值（x-1+绝对值x+1）*（绝对值y-2+绝对值y+1）*（绝对值z-3+绝对值z-1）=36 求x+2y+3z

Answer 1

|x-1+|x+1|| *(|y-2| + |y+1|)*(|z-3|+|z-1|)=36还是(|x-1| + |x+1| )*(|y-2| + |y+1|)*(|z-3|+|z-1|)=36如果是后者,我给出一个解如下：因为要求最大值,所以x,y,z最好是正数.假设,x>1,y>2,z>3 有（2x）*（2y-1）（2z-4）=36即x(2y-1)(3z-6)=27而x +2y -1 +3z-6 >=3(x(2y-1)(3z-6))^1/3=3*27^1/3=9从而x+2y+3z=x +2y -1 +3z-6+7 >=9+7 =16