(1)如果是1/(x²+4x+3)的话
1/(x²+4x+3)=1/(x+3)(x+1)=1/2 [1/(x+1)-1/(x+3)]
1/(x+1)=1/2-(1/2)^2(x-1)+(1/2)^3(1/2!)(x-1)^2+……+(-1)^n*(1/2)^(n+1) /n! (x-1)^n+........
1/(x+3)=1/4-(1/4)^2(x-1)+(1/4)^3(1/2!)(x-1)^2+……+(-1)^n*(1/4)^(n+1) /n! (x-1)^n+........
(2)如果是1/x²+4x+3的话
y=1/x展开为
1/x=1-(x-1)+1/2!(x-1)^2+……+(-1)^n/n! (x-1)^n+........
所以1/x²=[1-(x-1)+1/2!(x-1)^2+……+(-1)^n/n! (x-1)^n+........]²
=1-2(x-1)+(1/2+1/2+1)(x-1)^2+……
=1-2(x-1)+2(x-1)^2+……
所以1/x²+4x+3=1/x²+4(x-1)+7
=8+2(x-1)+2(x-1)^2+……
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