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y=(3x^2+5x+9)^5
y=( (3x+5)x+9 )^5
y=( (3x+5)x5+9 )^5+5乘( (3x+5)x5+9 )^4+10乘( (3x+5)x5+9 )^3+10乘( (3x+5)x5+9 )^2+5乘( (3x+5)x5+9 )^2+( (3x+5)x5+9 )
自己解埋落去-.- 多年冇返學 有錯咪屌

OHOHYA

原帖由 apoon123a 於 10-8-18 10:25 PM 發表

係第一條,
y= (3x-4)/(3x+4)
第二條,y=(3x^2+5x+9)^5
Ｙ＝０ 關鬼Factorization 事

Ｙ＝０ 同 Factorization  完全無關係
正如 y=f(x)=x^2+2x+1=(x+1)^2

人地條式 係用  Factorization  , ...

我鳩唱架咋,我都廢事諗

kyva1929

y= (3x-4)/(3x+4) * (3x+4)/(3x+4)

y = (3x-4)*(3x+4)
y = 9x^2-16

solving at y = 0, 自己 solve

[ 本帖最後由 kyva1929 於 10-8-18 10:23 AM 編輯 ]

bUＹ＊～

其中一個根為另一個根的負數
即係D>0 慢慢計

kyva1929

2) y=(3x^2+5x+9)^5

solving at y = 0

according to the fundamental theorems of algebra, this equation has ten roots, with which they're simply a repetition of the roots of the corresponding equation 3x^2+5x+9 = 0. (that is, each single complex root of this simple equation have itself repeating for 4 times). The complex root pair can be solved simply by using quadratic equation.

kyva1929

考慮二次方程 [(k+2)^2]+[(K-4)x] -4 =0，其中K≠-2 ，若其中一個根為另一個根的負數，求K的值。

我唔明咩叫其中一個根為另一個根的負數

let one of the root be A, the others be B = -A <--其中一個根為另一個根的負數
K^2+4K+(K-4)*A = 0  -(1)
K^2+4k+(K-4)*-A = 0 -(2)

solve the above equations simultaneously for K. Done.

hold on --
upon solving the above equations it gives nonsense. I bet you got the equation wrong.

[ 本帖最後由 kyva1929 於 10-8-18 10:38 AM 編輯 ]

CTPS

做乜學我喎..........

沙漠之狐~

根和根積

CTPS

學我問數仲要問D咁深...

你問D容易少少
比我威番次得唔得

bUＹ＊～

[(k+2)^2]+[(K-4)x] -4 =0
睇下有冇打錯