化简求值(1)若xm+2n=16,xn=2,(x≠0),求xm+n的值;(2)已知有理x满足x2-x+1=0,求(x-1)3+(x-1)2+(x-1)的值.
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化简求值 (1)若xm+2n=16,xn=2,(x≠0),求xm+n的值; (2)已知有理x满足x2-x+1=0,求(x-1)3+(x-1)2+(x-1)的值. |
答案
(1)∵xm+2n=16, ∴xm×(xn)2=16,∵xn=2, ∴xm×4=16, xm=4, ∴xm+n=xm×xn=4×2=8.
(2)(x-1)3+(x-1)2+(x-1) =(x-1)(x2+1-2x+x) =(x-1)(x2-x+1) ∵x2-x+1=0, ∴原式=(x-1)(x2-x+1)=0. |
举一反三
已知x2+2x-2=0,先将右式化简,再求值:(x-1)2+(x-3)(x+3)-(x-3)(x-1). |
计算:(2x+7)(3x-4)+(3x+5)(3-2x) |
先化简,再求值:(x-2)(x+2)-(x-2)2,其中x=-. |
先化简,再求值:(x+2y)2-2(x-y)(x+y)+2y(x-3y),其中x=-2,y=. |
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