(x-2)分之1+(x+2)分之1+(x平方+4)分之2x+(x四次方+16)分之4x三次方

来源:学生作业帮助网 编辑:作业帮 时间:2024/10/19 20:18:40
(x-2)分之1+(x+2)分之1+(x平方+4)分之2x+(x四次方+16)分之4x三次方
x]KPǿJ4).(]pe``EBd[اqg+Bm/AC"<yvRtQɠڕc@At`%E'Zv,se2Q;>V*m-uR$[rN͋0h,]W** \al 瘶.! ) !@92ʃ=٠OTzvg-\jԐ=ҤI7Y Ym"I],aXh5j&91$Ra!;(Z!Jf>U{Eݣ6B$n IMJLm_Di2parU6 }p:+^?_8"H(QJnK(FxLRu ߖ]D.]3H }/?5'6Ƒ;٪6ד1]@V]E)

(x-2)分之1+(x+2)分之1+(x平方+4)分之2x+(x四次方+16)分之4x三次方
(x-2)分之1+(x+2)分之1+(x平方+4)分之2x+(x四次方+16)分之4x三次方

(x-2)分之1+(x+2)分之1+(x平方+4)分之2x+(x四次方+16)分之4x三次方
=(x+2)/(x²-4)+(x-2)/(x²-4)+(x平方+4)分之2x+(x四次方+16)分之4x三次方
=2x/(x²-4)+(x平方+4)分之2x+(x四次方+16)分之4x三次方
=[2x(x²+4)+2x(x²-4)]/(x的4次方-16)+(x四次方+16)分之4x三次方
=4x³/(x的4次方-16)+4x³/(x的4次方+16)
=8x的7次方/(x的8次方-256)

关键点是二项二项通分,合并

1/(x-2)+1/(x+2)+2x/(x²+4)+4x^3/(x^4+16)
=【(x+2)+(x-2)】/【(x+2)(x-2)】+2x/(x²+4)+4x^3/(x^4+16)
=2x/(x²-4)+2x/(x²+4)+4x^3/(x^4+16)
=【2x(x²+4)+2x(x²-4)】/【(x²-...

全部展开

1/(x-2)+1/(x+2)+2x/(x²+4)+4x^3/(x^4+16)
=【(x+2)+(x-2)】/【(x+2)(x-2)】+2x/(x²+4)+4x^3/(x^4+16)
=2x/(x²-4)+2x/(x²+4)+4x^3/(x^4+16)
=【2x(x²+4)+2x(x²-4)】/【(x²-4)(x²+4)】+4x^3/(x^4+16)
=4x^3/(x^4-16)+4x^3/(x^4+16)
=【4x^3*(x^4+16)+4x^3*(x^4-16)】/【(x^4-16)*(x^4+16)】
=8x^7/(x^8-256)

收起

1/(x-2)+1/(x+2)+2x/(x²+4)+4x³/(x^4+16)=(x+2+x-2)/(x-2)(x+2)+2x/(x²+4)+4x³/(x^4+16)(前两项先通分)
=2x/(x²-4)+2x/(x²+4)+4x³/(x^4+16)=[2x(x²+4)+2x(x²-4)]/(x...

全部展开

1/(x-2)+1/(x+2)+2x/(x²+4)+4x³/(x^4+16)=(x+2+x-2)/(x-2)(x+2)+2x/(x²+4)+4x³/(x^4+16)(前两项先通分)
=2x/(x²-4)+2x/(x²+4)+4x³/(x^4+16)=[2x(x²+4)+2x(x²-4)]/(x²-4)(x²+4)+4x³/(x^4+16)
=4x³/(x^4-16)+4x³/(x^4+16)=[4x³(x^4+16)+4x³(x^4-16)]/(x^4-16)(x^4+16)
=8x^7/(x^8-256)

希望对你有所帮助

收起