题目内容
已知x+
=4,不求x的值,求下列各式的值:
(1)x2+
(2)(x-
)2.
1 |
x |
(1)x2+
1 |
x2 |
(2)(x-
1 |
x |
分析:(1)转化为完全平方的形式解答;
(2)将(x-
)2通过配方转化为(x+
)2-4的形式再代入求值.
(2)将(x-
1 |
x |
1 |
x |
解答:解:(1)x2+
=x2+2x•
+
-2x•
=(x+
)2-2
=42-2
=14;
(2)(x-
)2
=x2-2x•
+
=x2+2x•
+
-4x•
=(x+
)2-4
=16-4
=12.
1 |
x2 |
1 |
x |
1 |
x2 |
1 |
x |
=(x+
1 |
x |
=42-2
=14;
(2)(x-
1 |
x |
=x2-2x•
1 |
x |
1 |
x2 |
=x2+2x•
1 |
x |
1 |
x2 |
1 |
x |
=(x+
1 |
x |
=16-4
=12.
点评:本题考查了分式的混合运算和完全平方公式,适当转化式子是解题的关键.
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