Question

已知x,y,z,满足根号（3x-2y-4）+根号（2x-7y+3）=根号（2x-4y-z）*根号（-x/2 +2y）求z

Answer 1

由题可知：

3x-2y-4>=0 ===>3x-2y>=4 ===>9x-6y>=12,

2x-7y+3>=0 ===>2x-7y>=-3 ===>8x-28y>=-12,

所以17x-34y>=0,

x-2y>=0,

又-x+2y>=0 ===>x-2y=0 ===>x-2y>=z/2,

所以z