mathes多項數...

2007-09-12 6:30 am
我唔明點計,寫埋步驟,唔該..
加減法
1.(3p-4q+r)-(5q+3r-8p)
2.(6-r+r^2)-(2r-5-r^2)
3.(2s^2-9s+7)-(8-9s+3s^2)
4.(7s+5-8s^2)-(6s-s^2+1)
5.(s^3-5s+4-s^2)-(2s+7s^2+s^3+11)
6.(-7-2x^2+x^3-x)-(8+2x^2+9x-x^3)
乘法
1.(x+3)(x-4)
2.(3x+2)(5x-1)
3.(5x-7y)(x-2y)
4.(9x-2y)(2x-y)
5.5(x+2)(x-5)

回答 (2)

2007-09-12 6:51 am
✔ 最佳答案
1.(3p-4q+r)-(5q+3r-8p)
=3p-4q+r-5q-3r+8p
=11p-2r-9q

2.(6-r+r^2)-(2r-5-r^2)
=6-r+r^2-2r+5+r^2
=2r^2-3r+1

3.(2s^2-9s+7)-(8-9s+3s^2)
=2s^2-9s+7-8+9s-3s^2
=-s^2-1

4.(7s+5-8s^2)-(6s-s^2+1)
=7s+5-8s^2-6s+s^2-1
=s+4-7s^2

5.(s^3-5s+4-s^2)-(2s+7s^2+s^3+11)
=s^2-5s+4-s^2-2s-7s^2-s^3-11
=-7s^2-7s-7-s^2

6.(-7-2x^2+x^3-x)-(8+2x^2+9x-x^3)
=-7-2x^2+x^3-x-8-2x^2-9x+x^3
=-15-4x^2+2x^3-10x

乘法
1.(x+3)(x-4)
=(x^2+3x)+(-4x-12)
=x^2-x-12

2.(3x+2)(5x-1)
=(15x^2+10x)+(-3x-2)
=15x^2+7x-2

3.(5x-7y)(x-2y)
=(5x^2-7xy)+(10xy+14y^2)
=5x^2+3xy+14y^2

4.(9x-2y)(2x-y)
=(18x^2-4xy)+(-9xy+2y^2)
=18x^2-13xy+2y^2

5.5(x+2)(x-5)
=(5x+10)(5x-25)
=25x^2+50x-125x- 250
=25x^2-75x-250
2007-09-12 7:16 am
加減法
1) 3p-4q+r-5q-3r+8p
=11p-9q-2r

2)6-r+r^2-2r+5+r^2
=11-3r+2r^2

3)2s^2-9s+7-8+9S-3s^2
=-s^2-1

4)7s+5-8s^2-6s+s^2-1
=s+4-6s^2

5)s^3-5s+4-s^2-2s-7s^2-s^3-11
=-7s+3-9s^2

6)-7-2x^2+x^3-x-8-2x^2-9x+x^3
=-15-4x^2+2x^3-10x

乘法
1)x^2-4x+3X-12
=x^2-x-12

2)15x^2-3x+10x-2
=15x^2+7x-2

3)5x^2-10xy-(7xy-14y^2)
=5x^2-10xy-7xy+14y^2
=5x^2-17xy+14y^2

4)18x^2-9xy-(4xy-2y^2)
=18x^2-9xy-4xy+2y^2
=18x^2-13xy+2y^2

5)5(x^2-5x+2x-10)
=5(x^2-3x-10)
=5x^2-15x-50


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