[1] x/x^2+3x+2 +2/x^2+2x+1+-1/3x+6[2] 1/a+b-2a-1/a^2-b^2+b/b-a[3]x/3x-3-2x-1/6(x-1)^2

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[1]x/x^2+3x+2+2/x^2+2x+1+-1/3x+6[2]1/a+b-2a-1/a^2-b^2+b/b-a[3]x/3x-3-2x-1/6(x-1)^2[1]x/x^2+3x+2+2/x^

[1] x/x^2+3x+2 +2/x^2+2x+1+-1/3x+6[2] 1/a+b-2a-1/a^2-b^2+b/b-a[3]x/3x-3-2x-1/6(x-1)^2
[1] x/x^2+3x+2 +2/x^2+2x+1+-1/3x+6
[2] 1/a+b-2a-1/a^2-b^2+b/b-a
[3]x/3x-3-2x-1/6(x-1)^2

[1] x/x^2+3x+2 +2/x^2+2x+1+-1/3x+6[2] 1/a+b-2a-1/a^2-b^2+b/b-a[3]x/3x-3-2x-1/6(x-1)^2
第一题:原式=x/[(x+1)(x+2)]+2/(x+1)²-1/[3(x+2)]=3x(x+1)/[3(x+1)²(x+2)]+6(x+2)/[3(x+1)²(x+2)]-(x+1)²/[3(x+1)²(x+2)]=(3x²+3x+6x+12-x²-2x-1)/[3(x+1)²(x+2)]=(2x²+7x+11)/[3(x+1)²(x+2)]
第二题:原式=1/(a+b)-b/(a-b)-(2a-1)/(a²-b²)=(a-b-ab-b²)/(a²-b²)-(2a-1)/(a²-b²)=(a-b-ab-b²-2a+1)/(a²-b²)=(1-a-b-ab-b²)/(a²-b²)
第三题:原式=x/[3(x-1)]-(2x-1)/[6(x-1)²]=2x(x-1)/[6(x-1)²]-(2x-1)/[6(x-1)²]=(2x²-2x-2x+1)/[6(x-1)²]=(2x²-4x+1)/[6(x-1)²]

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