已知(3x-2)的二次方加y-3的绝对值=0,求5(2x-y)-2(6x-2y+2)+(4x-3y+1/2)的值
因为 (3x-2)的二次方加y-3的绝对值=0
所以
3x-2=0 y-3=0
x=2/3 y=3
5(2x-y)-2(6x-2y+2)+(4x-3y+1/2)
=10x-5y-12x+4y-4+4x-3y+1/2
=2x--4y+1/2
=2*2/3-4*3+1/2
=4/3-12+1/2
=8/6+3/6-72/6
=-61/6化简最后一步加的是二分之九是不是-91/65(2x-y)-2(6x-2y+2)+(4x-3y+9/2)
=10x-5y-12x+4y-4+4x-3y+9/2
=2x--4y+9/2
=2*2/3-4*3+9/2
=4/3-12+9/2
=8/6+27/6-72/6
=-37/6