Решите уравнение: (x^2-5x)^2+10x^2-50x+24=0.

\[\left( x^{2} - 5x \right)^{2} + 10x^{2} - 50x + 24 =\]

\[= 0\]

\[\left( x^{2} - 5x \right)^{2} + 10 \cdot \left( x^{2} - 5x \right) + 24 = 0\]

\[t = x^{2} - 5x\]

\[t^{2} + 10t + 24 = 0\]

\[D = 10^{2} - 4 \cdot 1 \cdot 24 =\]

\[= 100 - 96 = 4;\ \ \ \sqrt{D} = 2.\]

\[t_{1} = \frac{- 10 + 2}{2} = \frac{- 8}{2} = - 4;\ \ \ \ \]

\[t_{2} = \frac{- 10 - 2}{2} = \frac{- 12}{2} = - 6\]

\[x^{2} - 5x = - 4\]

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

\[D = ( - 5)^{2} - 4 \cdot 1 \cdot 4 =\]

\[= 25 - 16 = 9;\ \ \ \sqrt{D} = 3.\]

\[x_{1} = \frac{5 + 3}{2} = \frac{8}{2} = 4;\ \ \ \ \ \ \]

\[x_{2} = \frac{5 - 3}{2} = \frac{2}{2} = 1\]

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

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

\[D = ( - 5)^{2} - 4 \cdot 1 \cdot 6 =\]

\[= 25 - 24 = 1;\ \ \ \sqrt{D} = 1.\]

\[x_{1} = \frac{5 + 1}{2} = \frac{6}{2} = 3;\ \ \ \ \ \]

\[\ x_{2} = \frac{5 - 1}{2} = \frac{4}{2} = 2\]

\[Ответ:4;1;3;2.\]