請列步驟~! factorize
1) _____1 _____ _ _____ 1 _______ + ____1________ _ _____3____
........ ....x(x+1) ..... ....... . (x+1)(x+2) ........ . (x+2)(x+3) ......... x(x+3)
2) 1 - ( a _ ___1_____) * ____(a-2)^2______
....... ...... (a-1) .................. . a^2 -a+1
maths~~factorize (急!)~~~~~~~~~~~~
2006-10-17 3:08 am
回答 (2)
2006-10-17 4:49 am
✔ 最佳答案
1) _____1 _____ _ _____ 1 _______ + ____1________ _ _____3____........ ....x(x+1) ..... ....... . (x+1)(x+2) ........ . (x+2)(x+3) ......... x(x+3)
(x+2)(x+3) - x(x+3) + x(x+1) - 3(x+1)(x+2)
= ------------------------------------------------------------
x(x+1)(x+2)(x+3)
(x+2)[(x+3)-3(x+1)] + x[(x+1)-(x+3)]
= ------------------------------------------------------------
x(x+1)(x+2)(x+3)
-2x(x+2)-2x
= -----------------------------
x(x+1)(x+2)(x+3)
-2x^2-6x
= -----------------------------
x(x+1)(x+2)(x+3)
-2x(x+3)
= -----------------------------
x(x+1)(x+2)(x+3)
-2
= ---------------------
(x+1)(x+2)
2) 1 - ( a _ ___1_____) * ____(a-2)^2______
....... ...... (a-1) .................. . a^2 -a+1
係咪即係咁呀?
1 (a-2)^2
1 - (a - ------------) x ----------------
(a-1) (a^2-a+1)
2006-10-17 3:23 am
1) [1/x(x+1)]-[1/(x+1)(x+2)]+[1/(x+2)(x+3)]-[3/x(x+3)]
=[(x+2-x)/(x+1)(x+2)x]+[(x-x-2)/(x+2)(x+3)x]
=[2/(x+1)(x+2)x]-[2/(x+2)(x+3)x]
=2(x+3)-2(x+1)/x(x+1)(x+2)(x+3)
=4/x(x+1)(x+2)(x+3)
2)第二條我就睇唔到你寫咩。Sorry!
=[(x+2-x)/(x+1)(x+2)x]+[(x-x-2)/(x+2)(x+3)x]
=[2/(x+1)(x+2)x]-[2/(x+2)(x+3)x]
=2(x+3)-2(x+1)/x(x+1)(x+2)(x+3)
=4/x(x+1)(x+2)(x+3)
2)第二條我就睇唔到你寫咩。Sorry!
參考: me
收錄日期: 2021-04-23 12:37:08
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