close
標題:
f.4 amath
發問:
a)Expand(1-y^2)^4 b)using (a),expand(1-y)^4(1+y)^5 in ascending powers of y
最佳解答:
a)Expand(1-y^2)^4 (1-y^2)^4 = 1 - 4y^2 + 6y^4 -4y^6 + y^8 圖片參考：http://hk.yimg.com/i/icon/16/41.gif b)using (a),expand(1-y)^4(1+y)^5 in ascending powers of y (1-y)^4(1+y)^5 = (1+y)[(1-y)(1+y)]^4 = (1+y)(1-y^2)^4 = (1+y)(1-4y^2+6y^4-4y^6+y^8) = 1-4y^2+6y^4-4y^6+y^8 + y-4y^3+6y^5-4y^7+y^9 = 1+y-4y^2-4y^3+6y^4+6y^5-4y^6-4y^7+y^8+y^9 圖片參考：http://hk.yimg.com/i/icon/16/41.gif
其他解答:
a) (1-y^2)^4 = 1 - 4y^2 + 6y^4 - 4y^6 + y^8 b) (1-y)^4(1+y)^5 = [(1-y)^4(1+y)^4](1+y) = [(1-y)(1+y)]^4 (1+y) =(1-y^2)^4 (1+y) = (1-4y^2+6y^4-4y^6+y^8)(1+y) = (1-4y^2+6y^4-4y^6+y^8)+y(1-4y^2+6y^4-4y^6+y^8) = 1+y-4y^2-4y^3+6y^4+6y^5-4y^6-4y^7+y^8+y^9
f.4 amath
發問:
a)Expand(1-y^2)^4 b)using (a),expand(1-y)^4(1+y)^5 in ascending powers of y
最佳解答:
a)Expand(1-y^2)^4 (1-y^2)^4 = 1 - 4y^2 + 6y^4 -4y^6 + y^8 圖片參考：http://hk.yimg.com/i/icon/16/41.gif b)using (a),expand(1-y)^4(1+y)^5 in ascending powers of y (1-y)^4(1+y)^5 = (1+y)[(1-y)(1+y)]^4 = (1+y)(1-y^2)^4 = (1+y)(1-4y^2+6y^4-4y^6+y^8) = 1-4y^2+6y^4-4y^6+y^8 + y-4y^3+6y^5-4y^7+y^9 = 1+y-4y^2-4y^3+6y^4+6y^5-4y^6-4y^7+y^8+y^9 圖片參考：http://hk.yimg.com/i/icon/16/41.gif
其他解答:
a) (1-y^2)^4 = 1 - 4y^2 + 6y^4 - 4y^6 + y^8 b) (1-y)^4(1+y)^5 = [(1-y)^4(1+y)^4](1+y) = [(1-y)(1+y)]^4 (1+y) =(1-y^2)^4 (1+y) = (1-4y^2+6y^4-4y^6+y^8)(1+y) = (1-4y^2+6y^4-4y^6+y^8)+y(1-4y^2+6y^4-4y^6+y^8) = 1+y-4y^2-4y^3+6y^4+6y^5-4y^6-4y^7+y^8+y^9
此文章來自奇摩知識+如有不便請留言告知
文章標籤
全站熱搜
留言列表
