Даны функции y=f(x) и y=g(x), где f(x)=2/x, а g(x)=1/x. При каких значениях аргумента выполняется равенство g(x^2-4)-f(x+2)=-1?

\[y = f(x);\ \ \ y = g(x);\ \ \]

\[где\ \ f(x) = \frac{2}{x};\ \ \ g(x) = \frac{1}{x}\]

\[g\left( x^{2} - 4 \right) - f(x + 2) = - 1\]

\[\frac{1}{x^{2} - 4} - \frac{2}{x + 2} = - 1\]

\[\frac{1}{(x - 2)(x + 2)} - \frac{2}{x + 2} = - 1\]

\[\frac{1 - 2 \cdot (x - 2) + \left( x^{2} - 4 \right)}{(x - 2)(x + 2)} = 0\]

\[\frac{1 - 2x + 4 + x^{2} - 4}{x^{2} - 4} = 0\]

\[\frac{x^{2} - 2x + 1}{x^{2} - 4} = 0\]

\[\frac{(x - 1)^{2}}{\left( x^{2} - 4 \right)} = 0\]

\[x - 1 = 0\ \ \ \ \ \ \ x^{2} - 4 \neq 0\]

\[x = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x^{2} \neq 4\]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \neq \pm 2\]

\[Ответ:1.\ \]