问题 解答题
先能明白(1)小题的解答过程,再解答第(2)小题,
(1)已知a2-3a+1=0,求a2+
1
a2
的值.
由a2-3a+1=0知a≠0,∴a-3+
1
a
=0,即a+
1
a
=3
∴a2+
1
a2
=(a+
1
a
)2
-2=7;
(2)已知:y2+3y-1=0,求
y4
y8-3y4+1
的值.
答案

由y2+3y-1=0,知y≠0,∴y+3-

1
y
=0,即
1
y
-y=3,

(

1
y
-y)2=
1
y2
+y2-2=9,即
1
y2
+y2=11,

(

1
y2
+y2)2=121,

1
y4
+y4=119,

y8-3y4+1
y4
=y4-3+
1
y4
=116,

y4
y8-3y4+1
=
1
116

判断题
单项选择题