用 $a$ 表示方程 $\dfrac{1}{(x-a)^2}=|x-a|$ 的根。 在同一坐标系中，画出方程为 $y=\dfrac{1}{(x-a)^2}$ 和 $y=|x-a|$ 的曲线草图，其中 $a>1$。 由此解不等式 $\dfrac{1}{(x-a)^2}>|x-a|$。

1. Rewrite the equation using the property that $(x-a)^2 = |x-a|^2$.

Since $(x-a)^2 = |x-a|^2$, the equation can be rewritten as:
$\frac{1}{|x-a|^2} = |x-a|$

2. Rearrange and solve for $|x-a|$.

$1 = |x-a|^3$
$|x-a| = 1$

3. Solve the linear modulus equation for $x$.

$x - a = 1 \quad \text{or} \quad x - a = -1$
$x = a + 1 \quad \text{or} \quad x = a - 1$

1. Calculate the roots of the equation from part (i) to find the intersection points of the two graphs.

From part (i), we solve $\dfrac{1}{(x-a)^2} = |x-a|$:
Multiplying both sides by $|x-a|^2$ (assuming $x \neq a$) gives:
$1 = |x-a|^3 \implies |x-a| = 1$
$x - a = 1 \quad \text{or} \quad x - a = -1$
$x = a+1 \quad \text{or} \quad x = a-1$
The $y$-coordinate for both points is $y = |(a \pm 1) - a| = 1$.
Thus, the intersection points are $(a-1, 1)$ and $(a+1, 1)$.

2. Identify the key features of both graphs to sketch them on the same axes.

Curve $y=\dfrac{1}{(x-a)^2}$:

• Vertical asymptote at $x=a$ and horizontal asymptote at $y=0$.
• $y$-intercept at $(0, \dfrac{1}{a^2})$.
• The curve lies entirely above the $x$-axis, approaching infinity as $x \to a$.

Graph $y=|x-a|$:

• V-shape with its minimum point (vertex) at $(a, 0)$.
• $y$-intercept at $(0, a)$.

Sketch details:
Since $a > 1$, we have $a > \dfrac{1}{a^2}$, so the V-shape intersects the $y$-axis higher than the curve. Because $a > 1 \implies a-1 > 0$, both intersection points $(a-1, 1)$ and $(a+1, 1)$ lie in the positive $x$-axis region. Draw these features clearly on the axes.

3. Use the sketched graphs to solve the given inequality.

The inequality $\dfrac{1}{(x-a)^2} > |x-a|$ requires the curve $y=\dfrac{1}{(x-a)^2}$ to be strictly above the V-shaped graph $y=|x-a|$.
From the sketch, the curve is above the V-shape between the two intersection points $x=a-1$ and $x=a+1$.
However, the curve is undefined at the vertical asymptote $x=a$, so this value must be excluded from the region.
Therefore, the solution is:
$a-1 < x < a \quad \text{or} \quad a < x < a+1$