(x+1)(x-1)-(x-1)^2+(2x+1)(x-2),其中x=负3分之2,化简求值
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(x+1)(x-1)-(x-1)^2+(2x+1)(x-2),其中x=负3分之2,化简求值(x+1)(x-1)-(x-1)^2+(2x+1)(x-2),其中x=负3分之2,化简求值(x+1)(x-1)
(x+1)(x-1)-(x-1)^2+(2x+1)(x-2),其中x=负3分之2,化简求值
(x+1)(x-1)-(x-1)^2+(2x+1)(x-2),其中x=负3分之2,化简求值
(x+1)(x-1)-(x-1)^2+(2x+1)(x-2),其中x=负3分之2,化简求值
Solution:
yes?
(x+1)(x-1)-(x-1)²+(2x+1)(x-2)
=(x-1)(x+1-x+1)+(2x²-3x-2)
=2x-2+2x²-3x-2
=2x²-x-4
当x=-2/3时
原式=2×(-2/3)²-(-2/3)-4
=8/9+2/3-4
=-8/3
(x+1)(x-1)-(x-1)^2+(2x+1)(x-2),其中x=负3分之2,化简求值
解，得：
== (x-1) [(x+1) -(x-1)]+(2x+1)(x-2)
==(x-1) (x+1 -x+1)+(2x+1)(x-2)
==2(x-1)+(2x+1)(x-2)
==2x-2+2x^2-4x+x-2
==2x^2-x-4
所以
2x^2-x-4
==2(-2/3)^2-(-2/3)-4
==(8/9)-(-6/9)-4
==(2/9)-4
==-22/9
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