f.2的因式分解
回答 (3)
2008-03-02 5:10 am
✔ 最佳答案
(x - 2) (x - 1)^3 - (x - 2)^3 (x - 1)= (x - 2)(x - 1)[(x - 1)^2 - (x - 2)^2] << 抽共同因子 (x - 2)(x - 1)
= (x - 2)(x - 1){[x - 1 - (x - 2)][x - 1 + (x - 2)] <<< 利用恆等式 (x - y)(x + y) = x^2 - y^2
= (x - 2)(x - 1)(x - 1 - x + 2)(x - 1 + x - 2)
= (x - 2)(x - 1)(1)(2x - 3)
= (2x - 3)(x - 2)(x - 1)
2008-03-02 1:43 am
( x - 2 )( x - 1 )^3 - ( x - 1 )( x - 2 )^3
= ( x - 2 )( x - 1 )[ ( x - 1 )^2 - ( x- 2 )^2 ]
= ( x - 2 )( x - 1 )[ ( x - 1 ) + ( x - 2 ) ][ ( x - 1 ) - ( x - 2 ) ]
= ( x - 2 )( x - 1 )( 2x - 3 )( x - 1 - x + 2 )
= ( x - 2 )( x - 1 )( 2x - 3 )
= ( x - 2 )( x - 1 )[ ( x - 1 )^2 - ( x- 2 )^2 ]
= ( x - 2 )( x - 1 )[ ( x - 1 ) + ( x - 2 ) ][ ( x - 1 ) - ( x - 2 ) ]
= ( x - 2 )( x - 1 )( 2x - 3 )( x - 1 - x + 2 )
= ( x - 2 )( x - 1 )( 2x - 3 )
2008-03-01 9:34 pm
( x - 2 )( x - 1 )^3 - ( x - 1 )( x - 2 )^3
= ( x - 2 )( x - 1 )[ ( x - 1 )^2 - ( x- 2 )^2 ]
= ( x - 2 )( x - 1 )[ ( x - 1 ) + ( x - 2 ) ][ ( x - 1 ) - ( x - 2 ) ]
= ( x - 2 )( x - 1 )( 2x - 3 )( x - 1 - x + 2 )
= ( x - 2 )( x - 1 )( 2x - 3 )
= ( x - 2 )( x - 1 )[ ( x - 1 )^2 - ( x- 2 )^2 ]
= ( x - 2 )( x - 1 )[ ( x - 1 ) + ( x - 2 ) ][ ( x - 1 ) - ( x - 2 ) ]
= ( x - 2 )( x - 1 )( 2x - 3 )( x - 1 - x + 2 )
= ( x - 2 )( x - 1 )( 2x - 3 )
參考: 自己
收錄日期: 2021-04-14 20:20:56
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