(1/x)+x=3,则x²/(x⁴+x²+1)=?
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(1/x)+x=3,则x²/(x⁴+x²+1)=?
(1/x)+x=3,则x²/(x⁴+x²+1)=?
(1/x)+x=3,则x²/(x⁴+x²+1)=?
x²/(x⁴+x²+1)
=1/(x^2+1+1/x^2)
=1/[(x+1/x)^2-1]
=1/8
(1/x)+x=3,
(1/x)^2+2+x^2=9
(1/x)^2+x^2=7
则x²/(x⁴+x²+1)
=1/(x²+1+1/x²)
=1/(7+1)
=1/8
(1/x)+x=3 两边平方得:
1/x²+2+x²=9 即:1/x²+x²=7
x²/(x⁴+x²+1)
=1/(x²+1+1/x²)
=1/(7+1)
=1/8
分子分母都除以x²,得到1/x²+1+1/x²。x²+1+1/x²=[(1/x)+x]²-1=8.所以答案1/8