Soit \( A=\left(\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right) \). Montrer que, \( \forall n \in \mathbb{N}, A^{n}=\left(\begin{array}{ll}1 & n \\ 0 & 1\end{array}\right) \)

\( \frac{\frac{8}{5 x}-\frac{1}{x}}{\frac{4}{5}+\frac{3}{x}} \) \( \frac{\frac{8}{5 x}-\frac{1}{x}}{\frac{4}{5}+\frac{3}{x}}=\square( \) Simplify your answer. \( ) \)

Reste \( 2 x^{2}-7 x-1 \) de la suma de \( x^{2}+9 x-4 \) y \( -5 x^{2}-7 x+10 \)

Subtract. \[ \frac{3 x}{x^{2}-36}-\frac{2}{x+6} \]

a) \( \quad x^{2}=49 \)

Simplify. \( \frac{\frac{3}{x+5}}{x-5}-\frac{3}{x^{2}-25} \) \( \frac{\frac{3}{x+5}}{\frac{3}{x-5}-\frac{3}{x^{2}-25}}=\square \) (Use integers or fractions for any numbers in the expression.)