1、解:(1)原式=(4x) ²+2•4x•1/2+(1/2)²
=16x²+4x+1/4.
(2)原式=(1/4m)²-2•1/4 m•2n+(2n) ²
=1/16m²-mn+4n².
2、解:(1)原式=(2m) ²- (3n) ²-4m²-9n².
(2)原式=(-3a) ²-(1/2b) ²=9a²-1/4b2.
(3)原式=(y-4x) (y+4x)=y²-(4x) ²=y²-16x².
(4)原式=x²-y²+y²-z²一(x²-z²)
=x²-y2+y2-z²-x²+z²=0.
3、解:(1)(y+3) ² (3-y) ²
=[(y+3)(3-y)] ²
=(9-y²)²
=81-18y²+y4.
(2)(2a+b+1)(2a+6-1)
=[(2a+6)+1][(2a+6)-1]
=(2a+6) ²-1²
=4a2 +4ab+b²-1.
(3)(a-2b-3) (a+2b+3)
=[(2b+3)][a+(2b+3)]
=a²-(2b+3) ²
=a²-4b²-12b-9.
4、解:原式=25y²-45y-25y²
=-5y-1
当y2/5时,原式=-5×2/5-1=-3
5、解:(1)x²+2•x•1/4+(1/4) ²-[x²-(1/4)²]=1/4
x²+1/2x+1/16-x²+1/16=1/4,
1/2x=1/4-1/16-1/16,
1/2x=1/8,
x=1/4
(2)x²-1- (x²+4x+4) =7,
x²-1-x²-4x-4=7,
-4x=12,
X=-3.
6、解:2(x²-16)<2x²+5x-4x-10,
2x²-32<2x²+x-10,
-x<-10+32,
-x<22,
x>-22.
7、(1)2x 3y 12xy(答案不唯一)
(2)2 3 5x(答案不唯一)
(3)9/4 3/2
8、±4x或4x4
10、解:因为a+b+c=5,所以(a+b+c)²=25,
即a²+b²+c²+2ab+2bc+2ac=25.
又因为a²+b²+c²=3,
所以2ab+ 2bc+2ac=22.
所以ab+bc+ac=11.
11、解:πr²-π(r-2) ²
=πr²-π(r²-4r+4)
= πr²+4πr-4π
=4πr- 4π.
答:这个圆的面积减少(47π-4π)cm².