(x+2)(x-2)+(x-2)²+(x-4)(x-1),其中x²-3x=1过程!

问题描述:

(x+2)(x-2)+(x-2)²+(x-4)(x-1),其中x²-3x=1
过程!

(x+2)(x-2)+(x-2)²+(x-4)(x-1)
=(x-2)(x+2+x-2)+(x-4)(x-1)
=2x(x-2)+x²-5x+4
=2x²-4x+x²-5x+4
=3x²-9x+4
=3(x²-3x)+4
=3+4
=7

原式=x²-4+x²-4x+4+x²-5x+4
=3x²-9x+4
=3(x²-3x)+4
=3*1+4
=7

(x+2)(x-2)+(x-2)²+(x-4)(x-1)
=(x²-4)+(x²-4x+4)+(x²-5x+4)
=x²-4+x²-4x+4+x²-5x+4
=3x²-9x+4
=3(x²-3x)+4
当x²-3x=1时
原式 = 3×1+4=7