Question

Calculand $\frac{4}{x-1} - \frac{3+x}{x-1}$ , unde x apartine R-{1}

Answer 1

[tex] \frac{4}{x-1} - \frac{3+x}{x-1} = \frac{4-(3+x)}{x-1} = \frac{4-3-x}{x-1}=\\ = \frac{1-x}{x-1} = \frac{-(x-1)}{x-1} =-1[/tex]

Answer 2

$\frac{4}{x-1}- \frac{3+x}{x-1}= \frac{4-(3+x)}{x-1}= \frac{4-3-x}{x-1}= \frac{1-x}{x-1}=- \frac{x-1}{x-1}=-1$