x/x²-x+1=6求x²/x⁴+x²+1重在过程
问题描述:
x/x²-x+1=6求x²/x⁴+x²+1
重在过程
答
36/13
x/x²-x+1=6,都除以x,得:1/(x-1+1/x)=6
x+1/x=7/6,两边平方得:x2+1+1/x2=13/36
x²/x⁴+x²+1=36/13
答
由x/x²-x+1=6得:1/x-x=5
x²/x⁴+x²+1
=1/x²+x²+1
=(1/x-x)²+2+1
=5²+3
=28