问题
解答题
(1)分解因式:x7+x5+1 (2)对任何正数t,证明:t4-t+
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答案
(1)x7+x5+1=x7+x6+x5-x6+1
=x5(x2+x+1)-(x3+1)(x3-1)
=(x2+x+1)[x5-(x-1)(x3+1)]
=(x2+x+1)(x5-x4+x3-x+1),
(2)t4-t+
=(t4-t2+1 2
)+(t2-t+1 4
) 1 4
=(t2-
)2+(t-1 2
)2≥0 1 2
因为(t2-
)2与(t-1 2
)2不可能同时为0,故等于不成立,因此有:t4-t+1 2
>0.1 2