Решение.

Уравнение касательной   [tex]\bf y-y_0=y'(x_0)(x-x_0)[/tex]   .

[tex]\bf y=\dfrac{2x}{x-2}-3\ \ ,\ \ x_0=1\\\\\\y_0=y(x_0)=y(1)=\dfrac{2\cdot 1}{1-2}-3=-2-3=-5\\\\\\y'(x)=\dfrac{2(x-2)-2x\cdot 1}{(x-2)^2}=\dfrac{-4}{(x-2)^2}\\\\\\y'(x_0)=y'(1)=-\dfrac{4}{(1-2)^2}=-4\\\\\\y+5=-4(x-1)\\\\\\\boxed{\bf \ y=-4x-1\ }[/tex]