一道因式分解題,幫幫忙,謝謝

2022-04-19 07:08:24 字數 1016 閱讀 4727

1樓:風重的回憶

x(x+1)(x-1)+xy(x-y)-y(y+1)(y-1)=x(x^2-1)+xy(x-y)-y(y^2-1)=x^3-x+xy(x-y)-y^3+y;

=x^3-y^3-(x-y)+xy(x-y)=(x-y)(x^2+xy+y^2)-(x-y)+xy(x-y)=(x-y)(x^2+xy+xy+y^2-1)=(x-y)((x+y)^2-1)

=(x-y)(x+y-1)(x+y+1)

2樓:小南vs仙子

x(x+1)(x-1)+xy(x-y)-y(y+1)(y-1)=x(x+1)(x-1)+xy(x+1-1-y)-y(y+1)(y-1)

=x(x+1)(x-1)+xy(x+1)—xy(y+1)-y(y+1)(y-1)

=x(x+1)(x-1+y)-y(y+1)(x+y-1)=(x2+x-y2-y)(x+y-1)

=(x-y)(x+y+1)(x+y-1)

3樓:匿名使用者

很容易!

^3表示3次方

^2表示平方

x(x+1)(x-1)+xy(x-y)-y(y+1)(y-1)=x^3-x+xy(x-y)-y^3+y

=x^3-y^3-(x-y)+xy(x-y)=(x-y)(x^2+xy+y^2)-(x-y)+xy(x-y)=(x-y)(x^2+xy+y^2-1+xy)=(x-y)((x+y)^2-1)

=(x-y)(x+y-1)(x+y+1)

呵呵,多記記公式就會的!

4樓:泱柑枸

x(x+1)(x-1)+xy(x-y)-y(y+1)(y-1)=x^3-x+xy(x-y)-y^3+y

=x^3-y^3-(x-y)+xy(x-y)=(x-y)(x^2+xy+y^2)-(x-y)+xy(x-y)=(x-y)(x^2+xy+y^2-1+xy)=(x-y)((x+y)^2-1)

=(x-y)(x+y-1)(x+y+1)

其中x^3表示x的3次方

x^2表示x的2次方

初中因式分解問題。大家幫幫忙,初中因式分解問題。大家幫幫忙。(第四天)

1 題目出錯了,應為x x 6x 11x 6 x 6x 9x 2x 6 x x 6x 9 2 x 3 x x 3 2 x 3 x 3 x 3x 2 x 3 x 1 x 2 2 將 x 1 和 x 4 x 2 和 x 3 組合相乘 x 1 x 2 x 3 x 4 24 x 5x 4 x 5x 6 24...

整式的乘法與除法和因式分解謝謝了,大神幫忙啊

1 0.5b 0.2 a b 0.4a 0.3b 0.2a 2 1.2 a b a 2 ab 2 2.2x 2 y 2 2x 2 y 2 2x 2 y 2 2x y 2x y 3.m n 4 2 3 題目有錯?應該是 x 2 6x 9除以 2x 6 x 3 這樣的話答案為 x 2 6x 9 2x 6...

一道分解因式完全平方的題

解 a b 2c 4a 6b 8c 21 a 4a 4 b 6b 9 2c 8c 8 0 a 2 b 3 2 c 2 0平方項恆非負,三平方項之和 0,三項分別 0a 2 0 a 2 b 3 0 b 3 c 2 0 c 2 三角形是等腰三角形。已知 abc的三條邊a,b,c滿足等式a b 2c 4a...