初一数学题，因式分解
\u521d\u4e00\u6570\u5b66\u9898\uff1a\u56e0\u5f0f\u5206\u89e3\u89e3\uff1a\uff081\uff09x²+1=2x
\uff08x-1\uff09²=0
x=1
\uff082\uff09(x+2\uff09\uff08x+4\uff09=x²+6x+8
\u5219a=6
\uff08x+1\uff09\uff08x+9\uff09=x²+10x+9
\u5219b=9
x²+6x+9=\uff08x+3\uff09²
\uff083\uff09x²+x+1/4=0
\uff08x+1/2\uff09²=0
x=-1/2
\uff084\uff09\uff08x²-14x+49\uff09/\uff08x²-49\uff09=\uff08x-7\uff09²/\uff08x+7\uff09\uff08x-7\uff09
=\uff08x-7\uff09/\uff08x+7\uff09
\uff085\uff09x(x+y)(x-y)-x(x+y)²=x(x+y)(x-y-x-y)
=-2xy(x+y)
=-2*\uff08-1/2\uff09=1
ab\uff08a+b\uff09² -\uff08a+b\uff09-1
\u9898\u6709\u95ee\u9898
a²+ab-2b²
=(a+2b)(a-b)
x²-4xy+4y²-3x+3y+2
\u9898\u6709\u95ee\u9898
=a(a^2-b^2)
=a(a-b)(a+b)
2.已知x+y=6,xy=8,试求出x^2+y^2+3/2xy的值
(x^2+y^2+3)/(2xy)
=(x^2+y^2+2xy-2xy+3)/(2xy)
=[(x+y)^2-2xy+3]/(2xy)
=[6^2-2*8+3]/2*8
=[36-16+3]/16
=23/16
3.9998^2+2×9998
=9998(9998+2)
=9998*10000
=99980000
4.已知多项式ax^2+bx+c分解因式的结果是(3x+1)(4x-3),则a=,b=,c=
(3x+1)(4x-3)
=12x^2-5x-3
a=12,b=-5,c=-3
请采纳答案，支持我一下。
利用平方差公式
a^2-b^2=(a+b)*(a-b)
进行因式分解
答案是63 65
解：2^48-1=(2^24+1)(2^24-1)
=(2^24+1)(2^12+1)(2^12-1)
=(2^24+1)(2^12+1)(2^6+1)(2^6-1)
=(2^24+1)(2^12+1)x64x63
由此可见，这两个数分别为63和64；
分解的2的24次方加1乘以65和63，所以是65和63
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