問maths 功課 （中2）數學高手幫幫我吧（15分）

方程：
(a):5(x+8)-6(7-x)=42
5x+40-42+6x=42
11x=42-40+42
11x=44
x=4

(b): 3(1-a)+5(2-5a)=17-20(a+1)
3-3a+10-25a=17-20a-20
3+10-17+20=-20a+3a+25a
16=8a
a=2

(c): 2s/5-1/4=1/10
2s/5=1/10+1/4
2s/5=7/20
2s=7/4
s=7/8

(d): a-7/10=7-a
a+a=7+7/10
2a=77/10
a=3.85

(e): a/8-1=2a-5/2
a-8=16a-20
20-8=16a-a
12=15a
a=4/5

(f): m-2(1-m)/5=8
5m-2(1-m)=40
5m-2+2m=40
7m=42
m=6

(g) : 5(w/3-4)-3-w/6=41/2
5w/3-20-3-w/6=41/2
10w-120-18-w=123
9w=261
w=29

(h) : 1/2[2-1/5(4-t)]=-2/5
5[2-1/5(4-t)]=-4
10-(4-t)=-4
10-4+t=-4
t=-10


恒等式： **=2次方
（a) : 3z(5x+2y)(5x-2y)
=3z[(5x)^2-(2y)^2]
=3z(25x^2-4y^2)
=75(x^2)z-12(y^2)z

(b) : 2(5x+4y)**
=2[(5x)^2+2(5x)(4y)+(4y)^2]
=2(25x^2+40xy+16y^2)
=50(x^2)+80xy+32(y^2)

（c) : 4a(-b-3c)**
=4a[(-b)^2-2(-b)(3c)+(3c)^2
=4a(b^2+6bc+9c^2)
=4ab^2+24abc+36ac^2

(d):[1+1/x]**
=1^2+2(1)(1/x)+(1/x)^2
=1+2/x+1/(x^2)

(e) : [x/6-3/y]**
=(x/6)^2-2(x/6)(3/y)+(3/y)^2
=(x^2)/36-x/y+9/(y^2)

(f) : [x-5/3]**
=x^2-2(x)(5/3)+(5/3)^2
=(x^2)-10x/3+25/9

x^2=x的二次方

a.x=4
b.a=2
c.s=7/8
d.a=3.85
e.a=12/7
f.m=6
g.w=29
h.t=1.2

Factorize
a. 75x"z-12y"z
b.50x"+80xy+32y"
c.4ab"+24abc+36ac"
d.1+(2/x)+(1/x")
e.x"/36-x/y+9/y"
f.x"-10x/3+25/9

x"= x*x