试题:
已知x2-3x+1=0,求:
(1)x2+
1
x2
      
(2)
x4
x8+1

答案:

我来补答
(1)由x2-3x+1=0,可知:x≠0,
∴x2+1=3x,两边同时除以x得x+
1
x
=3,
∴(x+
1
x
2=32,即x2+
1
x 2
+2=9,
∴x2+
1
x 2
=7;

(2)∵x2+
1
x 2
=7,
∴(x2+
1
x 2
2=72,
∴x4+
1
x 4
+2=49,
∴x4+
1
x 4
=47,
x4
x8+1
=
1
x 4+
1
x 4
=
1
47
 
 
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