1﹚
1/﹙2-x﹚²=1/﹙x-2﹚²=﹙x+2﹚/[﹙x-2﹚²﹙x+2﹚]
1/[﹙x+2﹚﹙x-2﹚]=﹙x-2﹚/[﹙x-2﹚²﹙x+2﹚]
2﹚
1/﹙x²-1﹚=1/[﹙x+1﹚﹙x-1﹚]=﹙x-2﹚/[﹙x+1﹚﹙x-1﹚﹙x-2﹚]
2x/﹙x²-3x+2﹚=2x/[﹙x-1﹚﹙x-2﹚]=2x﹙x+1﹚/[﹙x+1﹚﹙x-1﹚﹙x-2﹚]
1﹚
1/﹙2-x﹚²=1/﹙x-2﹚²=﹙x+2﹚/[﹙x-2﹚²﹙x+2﹚]
1/[﹙x+2﹚﹙x-2﹚]=﹙x-2﹚/[﹙x-2﹚²﹙x+2﹚]
2﹚
1/﹙x²-1﹚=1/[﹙x+1﹚﹙x-1﹚]=﹙x-2﹚/[﹙x+1﹚﹙x-1﹚﹙x-2﹚]
2x/﹙x²-3x+2﹚=2x/[﹙x-1﹚﹙x-2﹚]=2x﹙x+1﹚/[﹙x+1﹚﹙x-1﹚﹙x-2﹚]