例:已知xx2+1=14,求代数式x2+1x2的值.
解:∵xx2+1=14,
∴x2+1x=4,即x2x+1x=4,
∴x+1x=4,
∴x2+1x2=x2+1x2+2-2=(x+1x)2-2=16-2=14.
根据材料回答问题:
(1)若3x2x2-32=15,求23x-12x的值;
(2)在(1)条件下求49x2+14x2的值;
(3)已知xx2-x+1=15,求x2+1x2的值.
x
x
2
+
1
=
1
4
x
2
+
1
x
2
x
x
2
+
1
=
1
4
x
2
+
1
x
=
4
x
2
x
+
1
x
=
4
x
+
1
x
=
4
x
2
+
1
x
2
=
x
2
+
1
x
2
+
2
-
2
=
(
x
+
1
x
)
2
-
2
=
16
-
2
=
14
3
x
2
x
2
-
3
2
=
1
5
2
3
x
-
1
2
x
4
9
x
2
+
1
4
x
2
x
x
2
-
x
+
1
=
1
5
x
2
+
1
x
2
【答案】(1)5;
(2);
(3)34.
(2)
25
2
3
(3)34.
【解答】
【点评】
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发布:2024/8/6 8:0:9组卷:68引用:2难度:0.5