Q:因式分解(x-2)^3+(2-x)(x^2-4x+1)
A:3(x-2)
---------------------可否給我詳解--------------------
(x^2代表x的平方,網路上不能上下標時通常都這樣打)
2005-08-17 14:32:52 · 5 個解答 · 發問者 Anonymous in 科學 ➔ 數學
答案是這樣的:
原式= (x - 2) ^3 - (x-2) (x^2 - 4x + 1)
= (x - 2) [(x-2)^2 - x^2 + 4x - 1]
= (x - 2) ( x^2 - 4x + 4 - x^2 + 4x - 1)
= 3(x - 2)
----------------------------------------------------------------------------
這題的重點部份就是把 (x - 2)給提出來而已,
其他部分應該都沒問題吧!
2005-08-17 14:43:29 · answer #1 · answered by 小胖子_Johnny 2 · 0⤊ 0⤋
(x-2)^3+(2-x)(x^2-4x+1)
=(x-2)(x-2)^2-(x-2)(x^2-4x+1)
=(x-2)(x^2-4x+4)-(x-2)(x^2-4x+1)
=(x-2)(x^2-4x+4-x^2+4x-1)
=3(x-2)
2005-08-17 16:13:17 · answer #2 · answered by ? 7 · 0⤊ 0⤋
(x-2)3+(2-x)(x2-4x+1)=(X-2)(X2-4X+4)+2x2-8x+2-x3+4x2-X=X3-4X2+4X-2X2-8X-8-x3+6x2-X+2=3X-6=3(X-2)
2005-08-17 14:39:37 · answer #3 · answered by 時均 7 · 0⤊ 0⤋
(x-2)^3+(2-x)(x^2-4x+1)
=(x-2)^2*(x-2)+(2-x)*(x^2-4x+1)
=(x-2)^2*(x-2)+(-1)*(x-2)*(x^2-4x+1)
=(x-2)^2*(x-2)-(x-2)*(x^2-4x+1)
=(x-2)*[(x-2)^2-(x^-4x+1)]
=(x-2)*[(x^2-4x+4)-(x^2-4x+1)]
=(x-2)*[(x^2-4x+4)-x^2+4x-1]
=(x-2)*3
2005-08-17 14:38:07 · answer #4 · answered by Anonymous · 0⤊ 0⤋
(x-2)^3-(x-2)(x^2-4x+1)=(x-2)(x^2-4x+4-x^2+4x-1)=(x-2)3=3(x-2)
2005-08-17 14:36:17 · answer #5 · answered by Anonymous · 0⤊ 0⤋