
0, x(x-1)>0,\\x\in (-\infty,0)U(1,+\infty)\\5)y=\frac{\sqrt{x^2+x}}{x+4}\\" alt="1) y=\sqrt{x-1}\\x-1 \geq 0, x \geq 1, x\in [1,+\infty)\\2) y=\sqrt{x^2-1}\\x^2-1 \geq 0, (x-1)(x+1) \geq 0,\\x\in (-\infty,-1]U[1,+\infty)\\3) y=\frac{x^2-9}{x^2-4}\\x^2-4\ne 0,x_1\ne-2, x_2\ne 2\\x\in (-\infty, -2)U(-2,2)U(2,+\infty)\\4)y=\frac{1}{\sqrt{x^2-x}}\\ x^2-x> 0, x(x-1)>0,\\x\in (-\infty,0)U(1,+\infty)\\5)y=\frac{\sqrt{x^2+x}}{x+4}\\" align="absmiddle" class="latex-formula">

0,\to x\ne 0, x\in (-\infty,0)U(0,+\infty)\\7) y=|log_2x|\\x>0\to x\in (0,+\infty)\\8) y=\sqrt{2^x}}\\2^x \geq 0" alt="6)y=log_2|x|\\|x|>0,\to x\ne 0, x\in (-\infty,0)U(0,+\infty)\\7) y=|log_2x|\\x>0\to x\in (0,+\infty)\\8) y=\sqrt{2^x}}\\2^x \geq 0" align="absmiddle" class="latex-formula">
Но показательная функция всегда >0, поэтому

0, x\in (-\infty,+\infty) \\ 9) y=log_2tgx\\tgx>0\\\pi n 0, x\in (-\infty,+\infty) \\ 9) y=log_2tgx\\tgx>0\\\pi n