(a) Prove that (x-2) ( x+1) = x^2-x-2 is an identity .
(b) Using the result of (a) , prove that (y-3)y = (y-1)^2 - y - 1 is also an identity.
只欠(b) 唔識, 煩請高手代解答.
thanks!!
F2 數學題
2014-09-06 7:34 pm
回答 (4)
2014-09-06 9:10 pm
✔ 最佳答案
(b)(y-1)^2-y-1
=(y-1)^2-(y-1)-2
let x=y-1 and substitute to (y-1)^2-y-1, then
(y-1)^2-(y-1)-2
=x^2-x-2
=(x-2)(x+1) from (a)
replace y-1 to x
(y-1-2)(y-1+1)
=(y-3)y
2014-09-07 9:28 pm
參考: Victor 老師
2014-09-06 11:59 pm
(y-3)y = (y-1)^2 - y - 1
y^-3y =2(y^-2y+1) - y - 1
y^-3y =2y^-4y+2 - y - 1
y^-3y =2y^-5y+1
so it is not dentity
y^-3y =2(y^-2y+1) - y - 1
y^-3y =2y^-4y+2 - y - 1
y^-3y =2y^-5y+1
so it is not dentity
2014-09-06 7:58 pm
設y=x+1
(y-3)y
=(x+1 - 3)(x+1)
=(x-2)(x+1)
=x^2 - x - 2
x=y-1
(y-3)y
=x^2 - x - 2
=(y-1)² - (y-1) - 2
=(y-1)² - y - 1
(y-3)y
=(x+1 - 3)(x+1)
=(x-2)(x+1)
=x^2 - x - 2
x=y-1
(y-3)y
=x^2 - x - 2
=(y-1)² - (y-1) - 2
=(y-1)² - y - 1
收錄日期: 2021-04-15 16:23:51
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