Решите уравнение: (x^2-5x)/2-3=0.

\[\text{\ \ }\frac{x^{2} - 5x}{2} - 3 = 0\]

\[\frac{x^{2} - 5x}{2} = 3\]

\[x^{2} - 6x = 6\]

\[x^{2} - 5x - 6 = 0\]

\[x_{1} + x_{2} = 5\]

\[x_{1} \cdot x_{2} = - 6 \Longrightarrow x_{1} = 6\ \ и\ \ \ x_{2} = - 1.\]

\[Ответ:x = 6\ \ \ и\ \ \ x = - 1.\]

\[x^{4} - 29x^{2} + 100 = 0\]

\[Пусть\ \ t = x^{2};\ \ t \geq 0\]

\[t^{2} - 29t + 100 = 0\]

\[t_{1} + t_{2} = 29\]

\[t_{1} \cdot t_{2} = 100 \Longrightarrow t_{1} = 4\ \ и\ \ \ t_{2} = 25\]

\[x^{2} = 4\ \ \ \ \ \ \ \ \ \ \ x^{2} = 25\]

\[x = \pm 2\ \ \ \ \ \ \ \ \ \ x = \pm 5\]

\[Ответ:\ \ x = \pm 2\ \ и\ \ \ x = \pm 5.\ \]

\[\frac{3x^{2} + 7x - 6}{4 - 9x^{2}} = \frac{(3x - 2)(x + 3)}{(2 - 3x)(2 + 3x)} =\]

\[= - \frac{x + 3}{2 + 3x}\]

\[3x^{2} + 7x - 6 = 3 \cdot \left( x - \frac{2}{3} \right)(x + 3) =\]

\[= (3x - 2)(x + 3)\]

\[D = b^{2} - 4ac = 49 - 4 \cdot 3 \cdot ( - 6) = 49 +\]

\[+ 72 = 121\]

\[x_{1} = \frac{- 7 + 11}{6} = \frac{4}{6} = \frac{2}{3}\]

\[x_{2} = \frac{- 7 - 11}{6} = - \frac{18}{6} = - 3.\]

\[x^{2} - 26x + q = 0\ \ \ \ \ \ и\ \ \ \ x_{1} = 12\]

\[x_{1} + x_{2} = 26 \Longrightarrow x_{2} + 12 = 26 \Longrightarrow x_{2} = 14\]

\[x_{1} \cdot x_{2} = q \Longrightarrow 12 \cdot 14 = q \Longrightarrow q = 168\]

\[Ответ:\ x_{2} = 14\ \ и\ \ \ q = 168.\]

\[\ 21x - x^{2} = 0\]

\[x(21 - x) = 0\]

\[x = 0\ \ \ \ \ \ \ 21 - x = 0\]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = 21\]

\[Ответ:x = 0\ и\ \ x = 21.\]