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[不等式] 来自某教师群的绝对值不等式恒成立$|x^2-ax-a^2|\le m$

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kuing posted 2016-4-12 15:17 |Read mode
广州王锦锦(7794*****) 11:13:37
QQ图片20160412151145.png
这个题还有没其他解法呢?答案也没看明白
设 $p=\max\{\abs x,\abs a\}$, $q=\min\{\abs x,\abs a\}$,则
\begin{align*}
\abs{x^2-ax-a^2}&\leqslant \abs{x^2-a^2}+\abs{ax} \\
&=p^2-q^2+pq \\
&\leqslant 1-q^2+q \\
&=\frac54-\left( q-\frac12 \right)^2 \\
&\leqslant \frac54,
\end{align*}
当 $x=1$, $a=-1/2$ 时 $\abs{x^2-ax-a^2}=5/4$,所以 $\abs{x^2-ax-a^2}$ 的最大值为 $5/4$,所以 $m$ 的取值范围是 $[5/4,+\infty)$。

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敬畏数学 posted 2016-4-29 09:53
回复 1# kuing
管理员的解法太神奇啦!!!

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游客 posted 2016-4-29 13:11
未命名.PNG

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游客 posted 2016-4-29 13:30
未命名.PNG

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敬畏数学 posted 2016-4-29 17:16
回复 4# 游客
此法很简单啊!

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