Question

Решите пожалуйста 2.1​

Answer 1

$\frac{\big{3}}{\big{x-2}}-\frac{\big{x+2}}{\big{x^2-2x+1}}\cdot\frac{\big{3x-3}}{\big{x^2-4}}=\frac{\big{3}}{\big{x-2}}-\frac{\big{x+2}}{\big{(x-1)^2}}\cdot\frac{\big{3(x-1)}}{\big{(x+2)(x-2)}}=\frac{\big{3}}{\big{(x-2)}}-\frac{\big{3}}{\big{(x-2)(x-1)}}=\frac{\big{3}}{\big{(x-2)}}\cdot(1-\frac{\big{1}}{\big{(x-1)}})=\frac{\big{3}}{\big{(x-2)}}\cdot\frac{\big{(x-2)}}{\big{x-1}}=\boxed{\frac{\big{3}}{\big{(x-1)}}}$