4(2p 3q)²-(3p-q)²-搜答案-拍题作业网

首先你要清楚P+Q的元素的个数,注意剔除重复的.4-1=3,4-2=2,4-3=1;5-1=4,5-2=3,5-3=2;6-1=5,6-2=4,6-3=3;所以P+Q的元素为1,2,3,4,5其真子集

(2p+q)(8p-3q)-2p(2p+q)=(2p+q)(8p-3q-2p)=(2p+q)(6p-3q)=3(2p+q)(2p-q)

(p+2q)^2-2(p+2q)(p+3q)+(p+3q)^2=（p+2q-p-3q）²=(-q)²=q²

平方差公式=[2(2p+3q)+(3p-q)][2(2p+3q)-(3p-q)]=(4p+6q+3p-q)(4p+6q-3p+q)=(7p+5q)(p+7q)

4(2p+3q)^2 -(3p-q)^2

4(2p+3q)^2-(3p-q)^2=16p^2+48pq+36q^2-9p^2+6pq-q^2=7p^2+54pq+35q^2如果本题有什么不明白可以追问,另外发并点击我的头像向我求助,请谅解,

你这个已经是分解了除非是化简吧再问：恩，怎么做再答：解(3p+2q)(3p-2q)(9p^2+4q^2)=[(3p)^2-(2q)^2](9p^2+4q^2)=(9p^2-4q^2)(9p^2+4q^

=(4p+6q+3p-q)(4p+6q-3p+q)=(7p+5q)(p+7q)

原式=(4p+6q)^2-(9p-3q)^2=(4p+6q+9p-3q)(4p+6q-9p+3q)=(13p+3q)(9q-5p)

4（2p+3q)^2-9（3p-q)^2=[2(2p+3q)+3(3p-q)][2(2p+3q)-3(3p-q)]=(4p+6q+9p-3q)(4p+6q-9p+3q)=(13p+3q)(9q-5p)

[(p-q)³-2(q-p)²-2/3(q-p)]/[(p-q)/3]=3(p-q)²-6(p-q)+2(p-q)

4（2p+3q)^2-(3p-q)^2=[2(2p+3q)-(3p-q)][2(2p+3q)+(3p-q)]=(4p+6q-3p+q)(4p+6q+3p-q)=(p+7q)(7p+5q)

(p+2q)(2p-q)-(p+q)(p-q)

=(2p²-pq+4pq-2q²)-(p²-q²)=p²+3pq-q²

那么8△m=4*8-(8+m)/2=10所以m=36

4(2p+3q)^2 - (3p-q)^2

（7p+5q)(p+7q)

解原式=(p²-4pq+4q²)-2(p²+5pq+6q²)+(p²+6pq+9q²)=(p²-2p²+p²)

3p^2+14pq+8q^2

p-q可能的取值是3,2,1,4,5就是说P※Q={1,2,3,4,5}有5个元素.所以真子集个数为2^5-1=32-1=31个

P(+)Q中可以有2,1,33－1＝23－2＝14－1＝34－2＝3（重复）所以,套用真子集公式,2的n次方（n为元素个数,本题中有三个元素,所以n＝3）答案是8

6p{(p+q)(p+q)}-4q(p+q)

6p{(p+q)(p+q)}-4q(p+q)=2(p+q)[3p(p+q)-2q]=2(p+q)(3p²++3pq-2q)