The equation of the tangent to the curve $y=\frac{x-5}{(x-3)(x+2)}$ at the point where it cuts the x-axis is :

$14 y -x+5 = 0$

$y=\frac{x-5}{(x-3)(x+2)}$

point of cut at x axis

y = 0

so $0=\frac{x-5}{(x-3)(x+2)} \Rightarrow x=5$ point is $(5,0)$

m = slope of tangent = $\frac{dy}{dx} = \frac{d}{dx}(\frac{x-5}{(x-3)(x+2)})$

$=\frac{d}{d x}\left(\frac{x-5}{x^2-x-6}\right)$

$=\frac{\left(x^2-x-6\right) \frac{d}{d x}(x-5)-(x-5) \frac{d}{d x}\left(x^2-x-6\right)}{\left(x^2-x-6\right)^2}$

$\frac{d y}{d x} =\frac{\left(x^2-x-6\right)-(x-5)(2 x-1)}{\left(x^2-x-6\right)^2}$

$=\frac{\left(x^2-x-6\right)-\left(2 x^2-x-10 x+5\right)}{\left(x^2-x-6\right)^2}$

$\Rightarrow \frac{d y}{d x}=\frac{\left(x^2-x-6\right)-\left(2 x^2-11 x+5\right)}{\left(x^2-x-6\right)^2}$

$\frac{d y}{d x} =\frac{-x^2+10 x-11}{\left(x^2-x-6\right)^2}$

so at $x=\left.5 \quad \frac{d y}{d x}\right|_{(5,0)}= \frac{-(5)^2+10(5)-11}{\left((5)^2-(5)-6\right)^2}=\frac{1}{14}$

So equation of tangent with $m=\frac{1}{14}$ and passing through (5, 0) is

$(y-0) =\frac{1}{14}(x-5)$

$14 y =x-5$