Example 14 Solve the equation $x^{2}+\frac{9 x^{2}}{(x-3)^{2}}=16$.

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Example 14 Solve the equation $x^{2}+\frac{9 x^{2}}{(x-3)^{2}}=16$.
Let $\frac{3 x}{x-3}=y$, then $x y=3 x+3 y$.
Thus, we have $\left\{\begin{array}{l}x^{2}+y^{2}=16, \\ x y=3(x+y) .\end{array}\right.$
By adding (1) and (2) $\cdot 2$, we get $(x+y)^{2}-6(x+y)-16=0$,
Solving this, we get $x+y=8$ or $x+y=-2$.
Therefore, we have $\left\{\begin{array}{l}x+y=8, \\ x y=24,\end{array}\right.$ or $\left\{\begin{array}{l}x+y=-2, \\ x y=-6 .\end{array}\right.$
The first set has no solution, solving the second set gives $x_{1}=-1+\sqrt{7}, x_{2}=-1-\sqrt{7}$.
Upon verification, $x_{1}$ and $x_{2}$ are both roots of the original equation.
1. A. Reason: From $0^{\circ}\sin \alpha>0$, and from $90^{\circ}\cos \beta, \cos \beta(\cos \alpha)$ ins.
From (1) and (2), we know $(\sin \alpha)^{\cos }>(\cos \alpha)^{\cos \beta}>(\cos \alpha)^{\sin \theta}$.