恆等式
1.若3x^2-(a+2)x+3b=cx^2+5x-9是恆等式,求a,b及c的值
2.若(3x+7)^2-(2x-5)(x+4)≡ax^2+bx+c,求a,b,及c的值
因式分解
1.9-(2x-y)^2
F.2數學題
2008-02-03 5:34 pm
回答 (2)
2008-02-03 6:06 pm
✔ 最佳答案
3x^2-(a+2)x+3b=cx^2+5x-9by comparing both sides,
c=3
and,
-(a+2)=5 ---> a=-7
and,
3b=-9 ---> b=-3
so,a=-7,b=-3,c=3
(3x+7)^2-(2x-5)(x+4)≡ax^2+bx+c
LHS=9x^2+42x+49-(2x^2-5x+8x-20)=7x^2+39x+69
by comparing both sides,
a=7
and,
b=39
and,
c=69
so,a=7,b=39,c=69
9-(2x-y)^2
=3^2-(2x-y)^2
=(3+2x-y)(3-2x+y)
2008-02-04 10:24:48 補充：
下面o個個抄我!
參考: 自己!
2008-02-03 8:31 pm
3x^2-(a+2)x+3b=cx^2+5x-9
by comparing both sides,
c=3
and,
-(a+2)=5 ---> a=-7
and,
3b=-9 ---> b=-3
so,a=-7,b=-3,c=3
(3x+7)^2-(2x-5)(x+4)≡ax^2+bx+c
LHS=9x^2+42x+49-(2x^2-5x+8x-20)=7x^2+39x+69
by comparing both sides,
a=7
and,
b=39
and,
c=69
so,a=7,b=39,c=69
2.9-(2x-y)^2
=3^2-(2x-y)^2
=(3+(2x-y))(3-(2x-y)) because(a+b)(a-b)≡a^2-b^2
=(3+2x-y)(3-2x+y)
by comparing both sides,
c=3
and,
-(a+2)=5 ---> a=-7
and,
3b=-9 ---> b=-3
so,a=-7,b=-3,c=3
(3x+7)^2-(2x-5)(x+4)≡ax^2+bx+c
LHS=9x^2+42x+49-(2x^2-5x+8x-20)=7x^2+39x+69
by comparing both sides,
a=7
and,
b=39
and,
c=69
so,a=7,b=39,c=69
2.9-(2x-y)^2
=3^2-(2x-y)^2
=(3+(2x-y))(3-(2x-y)) because(a+b)(a-b)≡a^2-b^2
=(3+2x-y)(3-2x+y)
參考: me
收錄日期: 2021-04-14 18:40:06
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080203000051KK00537