Question

Помогите, пожалуйста, даю 100 баллов

Answer 1

$y=\dfrac{2x+3}{x^2-1}\; \; ,\; \; x_0=2\\\\\\y'=\dfrac{2(x^2-1)-2x(2x+3)}{(x^2-1)^2}=\dfrac{2x^2-2-4x^2-6x}{(x^2-1)^2}=\dfrac{-2\, (x^2+3x+1)}{(x^2-1)^2}\; ,\\\\\\y'(2)=\dfrac{-2\cdot (4+6+1)}{9}=-\dfrac{22}{9}\\\\y(2)=\dfrac{7}{3}\\\\\\\boxed {\; y=y(x_0)+y'(x_0)\cdot (x-x_0)\; }\\\\\\y=\dfrac{7}{3}-\dfrac{22}{9}\cdot (x-2)\\\\\\y=-\dfrac{22}{9}\, x+\dfrac{65}{9}$