中一數學題(展開)

2008-11-18 2:11 am
要寫過程
1. (x-3)(x 2次方 +4x-9)+(7x 2次方 -6x+11)=?

2. (x-3)(x 2次方 +4x-9)+(7x 2次方 -6x+11)=?

3. (3x-2) 2次方 -(x-2)(x 2次方 -2x+1)=?

4. (14x 6次方 -9x 4次方 +22x 2次方 -10x)+(12x 4次方 +10x 2次方 -5x)=?

5.(9y 4次方 +5y 3次方 -6y)-(2y 3次方 -3y 2次方 +2y)=?

回答 (3)

2008-11-18 8:27 am
✔ 最佳答案
1.
(x-3)(x 2次方 +4x-9)+(7x 2次方 -6x+11)
=(x-3)(x² +4x-9)+(7x² -6x+11)
=x(x² +4x-9)-3(x² +4x-9) +(7x² -6x+11)
=x^3 +4x² -9x -3x² -12x +27 +7x² -6x+11
=x^3 +8x² -27x +38

2.
(x-3)(x 2次方 +4x-9)+(7x 2次方 -6x+11)
= 同上


3.
(3x-2) 2次方 -(x-2)(x 2次方 -2x+1)
=(3x-2)² -(x-2)(x²-2x+1)
=9x² -12x +4 -x(x²-2x+1) +2(x²-2x+1)
=9x² -12x +4 -x^3+ 2x²-x +2x²-4x+2
=-x^3+13x² -16x +6



4.
(14x 6次方 -9x 4次方 +22x 2次方 -10x)+(12x 4次方 +10x 2次方 -5x)
=(14x^6-9x^4+22x²-10x)+(12x^4+10x²-5x)
=14x^6-9x^4+22x²-10x+12x^4+10x²-5x
=14x^6+3x^4+32x²-15x


5.
(9y 4次方 +5y 3次方 -6y)-(2y 3次方 -3y 2次方 +2y)
=(9y^4+5y^3-6y)-(2y^3 -3y²+2y)
=9y^4+5y^3-6y -2y^3+3y²-2y
=9y^4+3y^3+3y²-8y
參考: 會考數學拿c既人~
2008-11-18 3:40 am
1. (x-3)(x²+4x-9)+(7x²-6x+11)
=x(x²+4x-9)-3(x²+4x-9)+7x²-6x+11
=x³+4x²-9x-3x²-12x+27+7x²-6x+11
=x³+4x²-3x²+7x²-9x-12x-6x+27+11
=x³+8x²-27x+38

2. = 1.

3.(3x-2)²-(x-2)(x²-2x+1)
=(3x-2)(3x-2)-[x(x²-2x+1)-2(x²-2x+1)]
=3x(3x-2)-2(3x-2)-(x³-2x²+x-2x²+4x-2)
=9x²-6x-6x+4-(x³-4x²+5x-2)
=9x²-12x+4-x³+4x²-5x+2
=6-17x+13x²-x³

4. (14x^6-9x^4+22x²-10x)+(12x^4+10x²-5x)
=14x^6-9x^4+22x²-10x+12x^4+10x²-5x
=14x^6+3x^4+32x²-15x

5. (9y^4+5y³-6y)-(2y³-3y²+2y)
=9y^4+5y³-6y-2y³+3y²-2y
=9y^4+3y³+3y²-8y
2008-11-18 2:46 am
1. (x-3)(x²+4x-9)+(7x²-6x+11)
=x(x²+4x-9)-3(x²+4x-9)+7x²-6x+11
=x³+4x²-9x-3x²-12x+27+7x²-6x+11
=x³+4x²-3x²+7x²-9x-12x-6x+27+11
=x³+8x²-27x+38

2. = 1.

3.(3x-2)²-(x-2)(x²-2x+1)
=(3x-2)(3x-2)-[x(x²-2x+1)-2(x²-2x+1)]
=3x(3x-2)-2(3x-2)-(x³-2x²+x-2x²+4x-2)
=9x²-6x-6x+4-(x³-4x²+5x-2)
=9x²-12x+4-x³+4x²-5x+2
=6-17x+13x²-x³

4. (14x^6-9x^4+22x²-10x)+(12x^4+10x²-5x)
=14x^6-9x^4+22x²-10x+12x^4+10x²-5x
=14x^6+3x^4+32x²-15x

5. (9y^4+5y³-6y)-(2y³-3y²+2y)
=9y^4+5y³-6y-2y³+3y²-2y
=9y^4+3y³+3y²-8y
參考: 自己


收錄日期: 2021-04-29 16:19:09
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20081117000051KK01240

檢視 Wayback Machine 備份