化简1/(x²+2x-1)-2/(x²+2x+1)+1/(x²+2x+3)=?
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化简1/(x²+2x-1)-2/(x²+2x+1)+1/(x²+2x+3)=?
解:令x²+2x+1=u,则x²+2x-1=u-2,x²+2x+3=u+2,代入原式得:
原式=1/(u-2)-(2/u)+1/(u+2)=[u(u+2)-2(u-2)(u+2)+u(u-2)]/[u(u-2)(u+2)]
=[(u²+2u)-(2u²-8)+(u²-2u)]/[(u-2)u(u+2)]
=8/[(x²+2x-1)(x²+2x+1)(x²+2x+3)]
解:令x²+2x+1=u,则x²+2x-1=u-2,x²+2x+3=u+2,代入原式得:
原式=1/(u-2)-(2/u)+1/(u+2)=[u(u+2)-2(u-2)(u+2)+u(u-2)]/[u(u-2)(u+2)]
=[(u²+2u)-(2u²-8)+(u²-2u)]/[(u-2)u(u+2)]
=8/[(x²+2x-1)(x²+2x+1)(x²+2x+3)]
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