我老師做個題:
(x-2)(x-4)≡ax^2+bx+c
x(x-4)-2(x-4)≡ax^2+bx+c
x^2-4x-2x+8≡ax^2+bc+c
x^2-6x+8≡ax^2+bx+c
by comparing coeft of x^2
a=1
by comparing coeft of x
b=6
by comparing coeft of constant
c=8
咁喱2題應該點做?
2x-8≡ax+b
ax+5≡b-7x
恆等式 [唔識做]
2008-10-21 4:34 am
回答 (3)
2008-10-21 4:43 am
✔ 最佳答案
2x-8≡ax+bby comparing coeft of x
a=2//
by comparing coeft of constant
b=-8//
ax+5≡b-7x
by comparing coeft of x
a=-7//
by comparing coeft of constant
b=5//
2008-10-21 5:00 am
2x-8≡ax+b
a=2,b=-8
ax+5≡b-7x
a=-7,b=5
a=2,b=-8
ax+5≡b-7x
a=-7,b=5
2008-10-21 4:52 am
2x-8≡ax+b
L.H.S.=2x-8
∵L.H.S=R.H.S
∴When comparing like terms on L.H.S and R.H.S., we have a=2;b=-8//
ax+5≡b-7x
L.H.S.=ax+5
R.H.S.=b-7x
=-(-7x+b)
=7x-b
∵L.H.S=R.H.S
∴When comparing like terms on L.H.S and R.H.S., we have a=7;-b=5//
2008-10-20 20:52:46 補充：
呢個方法簡單D..
L.H.S.=2x-8
∵L.H.S=R.H.S
∴When comparing like terms on L.H.S and R.H.S., we have a=2;b=-8//
ax+5≡b-7x
L.H.S.=ax+5
R.H.S.=b-7x
=-(-7x+b)
=7x-b
∵L.H.S=R.H.S
∴When comparing like terms on L.H.S and R.H.S., we have a=7;-b=5//
2008-10-20 20:52:46 補充：
呢個方法簡單D..
參考: MATHS書
收錄日期: 2021-04-23 20:34:31
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20081020000051KK01780