![image](https://tex.z-dn.net/?f=1%29+y%3D%5Csqrt%7Bx-1%7D%5C%5Cx-1+%5Cgeq+0%2C+x+%5Cgeq+1%2C+x%5Cin+%5B1%2C%2B%5Cinfty%29%5C%5C2%29+y%3D%5Csqrt%7Bx%5E2-1%7D%5C%5Cx%5E2-1+%5Cgeq+0%2C+%28x-1%29%28x%2B1%29+%5Cgeq+0%2C%5C%5Cx%5Cin+%28-%5Cinfty%2C-1%5DU%5B1%2C%2B%5Cinfty%29%5C%5C3%29+y%3D%5Cfrac%7Bx%5E2-9%7D%7Bx%5E2-4%7D%5C%5Cx%5E2-4%5Cne+0%2Cx_1%5Cne-2%2C+x_2%5Cne+2%5C%5Cx%5Cin+%28-%5Cinfty%2C+-2%29U%28-2%2C2%29U%282%2C%2B%5Cinfty%29%5C%5C4%29y%3D%5Cfrac%7B1%7D%7B%5Csqrt%7Bx%5E2-x%7D%7D%5C%5C+x%5E2-x%3E+0%2C+x%28x-1%29%3E0%2C%5C%5Cx%5Cin+%28-%5Cinfty%2C0%29U%281%2C%2B%5Cinfty%29%5C%5C5%29y%3D%5Cfrac%7B%5Csqrt%7Bx%5E2%2Bx%7D%7D%7Bx%2B4%7D%5C%5C)
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">
![image](https://tex.z-dn.net/?f=6%29y%3Dlog_2%7Cx%7C%5C%5C%7Cx%7C%3E0%2C%5Cto+x%5Cne+0%2C+x%5Cin+%28-%5Cinfty%2C0%29U%280%2C%2B%5Cinfty%29%5C%5C7%29+y%3D%7Clog_2x%7C%5C%5Cx%3E0%5Cto+x%5Cin+%280%2C%2B%5Cinfty%29%5C%5C8%29+y%3D%5Csqrt%7B2%5Ex%7D%7D%5C%5C2%5Ex+%5Cgeq+0)
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
![image](https://tex.z-dn.net/?f=+2%5Ex%3E0%2C+x%5Cin+%28-%5Cinfty%2C%2B%5Cinfty%29+%5C%5C+9%29+y%3Dlog_2tgx%5C%5Ctgx%3E0%5C%5C%5Cpi+n+%3Cx%3C%5Cfrac%7B%5Cpi+%7D%7B2%7D%2B%5Cpi+n%2C+n%5Cin+Z)
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