1、
原式=1\(x+1)-[(x+3)\(x+1)(x-1)]*[(x-1)^2\(x+1)(x+3)]
=1\(x+1)-(x+3)(x-1)^2\[(x+1)^2(x-1)(x+3)]
=(x+1)\(x+1)^2-(x-1)\(x+1)^2
=(x+1-x+1)\(x+1)^2
=2\(x+1)^2
=2\(2+1)^2
=2\9
2、
3x+y=24 (1)
4x+ay=18 (2)
(1)*4-(2)*3
12x+4y-12x-3ay=96-54
(4-3a)y=42
y=42/(4-3a)是正整數(shù)
所以4-3a是42的正約數(shù)
所以4-3a=1,2,3,6,7,14,21,42
且a是整數(shù)
所以4-3a=1,a=1,則y=42/(4-3a)=42
4-3a=7,a=-1,則y=42/(4-3a)=6
把a=1和a=-1代入4x+ay=18
a=1,y=42,4x+42=18,x不是正整數(shù)
a=-1,y=6,4x-6=18,x=6,是正整數(shù)
所以a=-1