已知 $a$ 是正常数，解不等式 $|x-3a|>|x-a|$ 。

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完整解法

1. Eliminate the modulus by squaring both sides of the inequality.

$|x-3a| > |x-a|$
$(x-3a)^2 > (x-a)^2$

2. Expand the squared expressions.

$x^2 - 6ax + 9a^2 > x^2 - 2ax + a^2$

3. Simplify the inequality by grouping like terms.

$9a^2 - a^2 > x^2 - x^2 - 2ax + 6ax$
$8a^2 > 4ax$
$2a^2 > ax$

4. Solve for $x$, using the given condition that $a$ is positive.

Since $a$ is a positive constant ( $a > 0$ ), dividing both sides by $a$ preserves the inequality sign:
$2a > x$
$x < 2a$

答案

$x < 2a$

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