Найдите область определения функции: y=√(x^2+4x-5)/(3x-4).

Ответ:

\[y = \frac{\sqrt{x^{2} + 4x - 5}}{3x - 4}\ \]

\[3x - 4 \neq 0\]

\[3x \neq 4\]

\[x \neq \frac{4}{3} \neq 1\frac{1}{3}.\]

\[x^{2} + 4x - 5 \geq 0\]

\[D = 4 + 5 = 9\]

\[x_{1} = - 2 + 3 = 1;\]

\[x_{2} = - 2 - 3 = - 5\]

\[(x + 5)(x - 1) \geq 0\]

\[Но\ x \neq 1\frac{1}{3}:\]

\[x \in ( - \infty; - 5\rbrack \cup \left\lbrack 1;1\frac{1}{3} \right) \cup \left( 1\frac{1}{3}; + \infty \right).\]