已知x/xˆ2-x+1 =7,求xˆ2/xˆ4+xˆ2+1(用倒数法计算最好)!

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/20 22:27:25
已知x/xˆ2-x+1 =7,求xˆ2/xˆ4+xˆ2+1(用倒数法计算最好)!
xN@_Q2xl vX@[ NU*V\ZD$ };W`Ġٰ`V̯?^~ {)H?Lz#'Jf^):zzvob?vޟǽ)mq{WcOe' 咢u#T읲:U$M,tDy֒CNMˊt|a$1h.M;\io׿2N%~'C6a[8!a0R}i/cud: VG$Ƣ _IעMI5%~-v9FaIFswv _hC8c|t I]ˠԱ l; ;- Z aD:-jMKM,U k c&,a$\r\ zg H

已知x/xˆ2-x+1 =7,求xˆ2/xˆ4+xˆ2+1(用倒数法计算最好)!
已知x/xˆ2-x+1 =7,求xˆ2/xˆ4+xˆ2+1(用倒数法计算最好)!

已知x/xˆ2-x+1 =7,求xˆ2/xˆ4+xˆ2+1(用倒数法计算最好)!
x/(xˆ2-x+1 )=7 吗?
x-1+1/x=1/7 x+1/x=8/7
xˆ2/(xˆ4+xˆ2+1)=1/(x²+1+1/x²)=1/[(x+1/x)²+1] = 1/[(8/7)²+1]=49/113

ghjgh的崇高的吧打发

x/xˆ2-x+1 =7
1/(x-1+(1/x))=7
x+(1/x)=8/7
xˆ2/xˆ4+xˆ2+1
=1/(x²+1+(1/x²))
=1/((x+(1/x))²-1)