16..中四MATH..請詳列步驟..

設g(x)=6x^4 - 2cx + 1
a)求g(d)和g(-d)

g(d)=6d^4 - 2cd + 1
g(-d)=6(-d)^4- 2c(-d) + 1
=6d^4 + 2cd + 1

b)若對於任何實數,d,g(d)=g(-d),求b的值
g(d)=g(-d)
6d^4 - 2cd + 1= 6d^4 + 2cd + 1
6d^4 - 6d^4 - 2cd + 1= 2cd + 1
-2cd + 1=2cd + 1
2cd + 2cd=0
4cd=0
cd=0
c=0
Put c=0 into g(x)=6x^4 - 2cx + 1
g(x)=6x^4+1
g(-1)=6(-1)^4+1
g(-1)=6+1
g(-1)=7

g(d)=6d^4-2cd+1

g(-d)=6(-d)^4-2c(-d)+1
=6d^4-2cd+1

b)唔明b點出黎...

a) g(d) = 6d^4-2cd+1

g(-d) = 6d^4+2cd+1

g(d) = g(-d)
6d^4-2cd+1 = 6d^4+2cd+1
-2cd=2cd
2cd+2cd=0
4cd=0

但係 b 的值.........unknown

A)g(d)=6d^4-2cd+1
g(-d)=6(-d)^4-2c(-d)+1
=6d^4+2cd+1
b) g(d)=g(-d)
6d^4-2cd+1=6d^4+2cd+1
6d^4-6d^4-2cd+1=2cd+1
-2cd+1=2cd+1
- 2cd - 2cd=0
之後自己計

a) g(d) = 6d^4-2cd+1
g(-d) = 6(-d)^4-2c(-d)+1
= 6d^4+2cd+1
b) =.=*我穩唔到邊度有b......

希望我係咁而幫到下你啦!!!