Question

Help me with this please

Answer 1

Answer:

$\large\boxed{(b)\ \dfrac{x+2}{x-3}}$

Step-by-step explanation:

$\dfrac{1}{x+1}+\dfrac{x}{x-3}-\dfrac{-x-5}{x^2-2x-3}=(*)\\\\x^2-2x-3=x^2-3x+x-3=x(x-3)+1(x-3)=(x-3)(x+1)\\\\(*)=\dfrac{1(x-3)}{(x+1)(x-3)}+\dfrac{x(x+1)}{(x+1)(x-3)}+\dfrac{-(-x-5)}{(x+1)(x-3)}\\\\=\dfrac{x-3+x^2+x+x+5}{(x+1)(x-3)}=\dfrac{x^2+(x+x+x)+(-3+5)}{(x+1)(x-3)}\\\\=\dfrac{x^2+3x+2}{(x+1)(x-3)}=\dfrac{x^2+2x+x+2}{(x+1)(x-3)}=\dfrac{x(x+2)+1(x+2)}{(x+1)(x-3)}\\\\=\dfrac{(x+2)(x+1)}{(x+1)(x-3)}\qquad\text{cancel (x + 1)}\\\\=\dfrac{x+2}{x-3}$