![y = \dfrac{\sqrt{3 - x}}{3^{x} - 1}](https://tex.z-dn.net/?f=y+%3D+%5Cdfrac%7B%5Csqrt%7B3+-+x%7D%7D%7B3%5E%7Bx%7D+-+1%7D)
Чтобы найти область определения функции, мы должны учесть два условия (ОДЗ):
![\left \{ {\bigg{3 - x \geqslant 0 \ } \atop \bigg{3^{x} - 1 \neq 0}} \right. \ \ \ \ \ \ \ \ \ \ \ \ \ \left \{ {\bigg{x \leqslant 3 \ } \atop \bigg{3^{x} \neq 1}} \right.\\\\\left \{ {\bigg{x \leqslant 3 \ \ \ } \atop \bigg{3^{x} \neq 3^{0}}} \right. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left \{ {\bigg{x \leqslant 3} \atop \bigg{x \neq 0}} \right.](https://tex.z-dn.net/?f=%5Cleft+%5C%7B+%7B%5Cbigg%7B3+-+x+%5Cgeqslant+0+%5C+%7D+%5Catop+%5Cbigg%7B3%5E%7Bx%7D+-+1+%5Cneq+0%7D%7D+%5Cright.+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5Cleft+%5C%7B+%7B%5Cbigg%7Bx+%5Cleqslant+3+%5C+%7D+%5Catop+%5Cbigg%7B3%5E%7Bx%7D+%5Cneq+1%7D%7D+%5Cright.%5C%5C%5C%5C%5Cleft+%5C%7B+%7B%5Cbigg%7Bx+%5Cleqslant+3+%5C+%5C+%5C+%7D+%5Catop+%5Cbigg%7B3%5E%7Bx%7D+%5Cneq+3%5E%7B0%7D%7D%7D+%5Cright.+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5Cleft+%5C%7B+%7B%5Cbigg%7Bx+%5Cleqslant+3%7D+%5Catop+%5Cbigg%7Bx+%5Cneq+0%7D%7D+%5Cright.)
Итак, объединяем оба условия и получаем: ![x \in (-\infty; 0) \cup (0; \ 3]](https://tex.z-dn.net/?f=x+%5Cin+%28-%5Cinfty%3B+0%29+%5Ccup+%280%3B+%5C+3%5D)
Ответ: ![D(y): \ x \in (-\infty; 0) \cup (0; \ 3]](https://tex.z-dn.net/?f=D%28y%29%3A+%5C+x+%5Cin+%28-%5Cinfty%3B+0%29+%5Ccup+%280%3B+%5C+3%5D)
Кос а= 1 -16\25= 9\25= 3\5
тг а= -4\5 * 5/3= -4\3
аналогично котангенс а = -3/4
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