1)

2)

3)

4)
0\qquad \\2x+2>0\qquad \end{matrix}\quad \begin{Bmatrix}x(x-3)+(x-3)=0\\|x|>1\qquad \qquad \qquad \\x>-1\qquad \qquad \qquad \end{matrix}" alt="\displaystyle \log_5(x^2-1)=\log_5(2x+2)\\\\\begin{Bmatrix}x^2-1=2x+2\\x^2-1>0\qquad \\2x+2>0\qquad \end{matrix}\quad \begin{Bmatrix}x(x-3)+(x-3)=0\\|x|>1\qquad \qquad \qquad \\x>-1\qquad \qquad \qquad \end{matrix}" align="absmiddle" class="latex-formula">
(x-3)(x+1) = 0 ⇔ x=3 или x=-1
x=3 подходит под условие |x|>1 и x>-1
x=-1 не подходит под условие x>-1.
Ответ: 3.