Решите уравнение методом замены переменной: 21/(x^2-4x+10)-x^2+4x=6.

\[\frac{21}{x^{2}x + 10} - x^{2} + 4x = 6\]

\[Пусть\ \ t = x^{2} - 4x + 10:\]

\[\frac{21}{t} - (t - 4) = 0\ \ \ | \cdot t\]

\[21 - t(t - 4) = 0\]

\[21 - t^{2} + 4t = 0\]

\[t² - 4t - 21 = 0\]

\[t_{1} + t_{2} = 4;\ \ t_{1} \cdot t_{2} = - 21\]

\[t_{1} = 7,\ \ t_{2} = - 3.\]