Question

已知(3x-2)的2次方+2x-y-3的绝对值=0，求5（2x-y）-2(6x-2y+2）+（4x-3y-1/2)的值 急，求过程答案，谢谢

Answer 1

(3x-2)的2次方+2x-y-3的绝对值=0
∴﹛3x-2=0, 2x-y-3=0
x=2/3, y=-5/3
5（2x-y）-2(6x-2y+2）+（4x-3y-1/2)
=10x-5y-12x+4y-4+4x-3y-½
=2x-4y-4.5
=4/3+20/3-4.5
=8-4.5
=3.5

Answer 2

因为(3x-2)²+l2x-y-3l=0,即3x-2=0,2x-y-3=0，所以x=2/3,y=-5/3,
5（2x-y）-2(6x-2y+2）+（4x-3y-1/2)
=10x-5y-12x+4y-4+4x-3y-1/2
=2x-4y-9/2
=2X2/3-4X(-5/3)-9/2
=4/3+20/3-9/2
=8-9/2
=7/2
祝你学习进步！

Answer 3

由已知条件知，3x-2=0,2x-y-3=0,所以x=2/3,y=-5/3.
原式=2x-4y-9/2=7/2

Answer 4

（3x-2）^2+|2x-y-3|=0可以得出x=2/3，y=-4/3，然后把x，y代入后面的式子即可