數學問題一問,數學高手請進...thx!

因式分解
1. x^2+7x+10= (x + 2)(x + 5)2. x^2-6x+5= (x - 1)(x - 5)3. x^3+64= x^3 + 4^3= (x + 4)(x^2 - 4x + 4^2)= (x + 4)(x^2 - 4x + 16)
展開
1. (pq-rs)^2= (pq)^2 - 2 pq rs + (rs)^2= (p^2)q^2 - 2pq + (r^2)(s^2)2. (x-y+z)^2= (x-y)^2 + 2(x-y)z + z^2= x^2 - 2xy + y^2 + 2xz - 2yz + z^2= x^2 + y^2 + z^2 + 2xz - 2xy - 2yz
求以下兩個含x的恆等式中的A和B
1. (A-B)x+(A+B)≡7x+3比較係數 :A-B = 7....(1)
A+B = 3....(2)(1)+(2) :2A = 7+3A = 5 , 代入(1) : 5 - B = 7
B = - 22. x^2+AX+10≡(x-2)(x+B)x^2+Ax+10 ≡ x^2 + (B-2)x - 2B比較係數 :A = B-2 ....(1)
10 = - 2B ....(2)由(2) , B = - 5 , 代入(1) :A = -5-2 = - 7另法 , 當 x = 0 ,
10 = - 2B , B = - 5 ,
當 x = 1 ,
1 + A + 10 = 1 + (B-2) - 2B A + 11 = 1 + (-5-2) - 2(-5)A = - 7 求一凸7邊形的內角和(7 - 2) x 180 = 900°
若一正多邊形的一內角是156度,求其邊數設其邊數 = n , (n - 2) x 180 = 156n180n - 360 = 156n24n = 360n = 15邊數 = 15

Sorry!!!!

x^2+7x+10
(x+5)(x+2)
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x^2-6x+5
(x-1)(x-5)
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x^3+64
(x)^3-4^3
(x-3)(x^2+3x+9)
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(pq-rs)^2
p^2q^2-2pqrs+r^2s^2
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(x-y+z)^2
((x-y)+z)^2
(x-y)^2+2z(x-y)+z^2
x^2+2xy-y^2+2xz-2yz+z^2
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(A-B)x+(A+B)≡7x+3
....................
A-B=7
A...=7+B
............
A+B=3
(7+B)+B=3
2B+7.....=3
2B.........=-4
B...........=-2

A=7+(-2)=5
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x^2+Ax+10≡(x-2)(x+B)


LHS:x^2+Ax+10


RHS=(x-2)(x+B).
.......=x(x+B)-2(x+B)
.......=x^2+Bx-2x-2B
.......=x^2+x(B-2)-2B

............................
B=-5

-5-2=A
A...=-7
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求一凸7邊形的內角和
(7-2)x180
=900degrees
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若一正多邊形的一內角是156度,求其邊數
(n-2)x180=156n
180n-360.=156n
24n.........=360
n.............=15
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