题目内容
已知x+
=2,则下列等式成立的有( )
①x2+
=2;②x4+
=2;③x8+
=2;④x-
=0.
1 |
x |
①x2+
1 |
x2 |
1 |
x4 |
1 |
x8 |
1 |
x |
分析:根据完全平方公式求出x2+
=(x+
)2-2•x•
=22-2=2,x4+
=(x2+
)2-2•x2•
=22-2=2,x8+
=(x4+
)2-2=2,(x-
)2=(x+
)2-4•x•
=22-4=0,判断即可.
1 |
x2 |
1 |
x |
1 |
x |
1 |
x4 |
1 |
x2 |
1 |
x2 |
1 |
x8 |
1 |
x4 |
1 |
x |
1 |
x |
1 |
x |
解答:解:∵x+
=2,
∴x2+
=(x+
)2-2•x•
=22-2=2,∴①正确;
x4+
=(x2+
)2-2•x2•
=22-2=2,∴②正确;
x8+
=(x4+
)2-2=2,∴③正确
(x-
)2=(x+
)2-4•x•
=22-4=0,∴④正确;
故选D.
1 |
x |
∴x2+
1 |
x2 |
1 |
x |
1 |
x |
x4+
1 |
x4 |
1 |
x2 |
1 |
x2 |
x8+
1 |
x8 |
1 |
x4 |
(x-
1 |
x |
1 |
x |
1 |
x |
故选D.
点评:本题考查了完全平方公式的应用,注意:(a+b)2=a2+2ab+b2,(a-b)2=a2-2ab+b2.

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