2x2-[x2-2(x2-3x-1)-3(x2-1-2x)]其中:x=1/2.

问题描述:

2x2-[x2-2(x2-3x-1)-3(x2-1-2x)]其中:x=

1
2

原式=2x2-(x2-2x2+6x+2-3x2+3+6x)
=2x2-(-4x2+12x+5)
=6x2-12x-5
∵x=

1
2

代入原式可得:6×
1
4
-12×
1
2
-5=-
19
2