Муодилаҳоро ҳал кунед:

\[C_x^1 + 6C_x^2 + 6C_x^3 = 9x^2 - 14x.\]

Ҳал.

\(C_n^m = \frac{n!}{(n - m)! m!}\), ки дар ин ҷо \(m \leq n;\quad C_n^0 = 1\)

\(C_x^1 = \frac{x!}{(x - 1)! \cdot 1!} = \frac{x \cdot (x - 1)!}{(x - 1)! \cdot 1} = \frac{x}{1} = x\)

\(C_x^2 = \frac{x!}{(x - 2)! \cdot 2!} = \frac{x \cdot (x - 1) \cdot (x-2)!}{(x - 2)! \cdot 2} = \frac{x(x - 1)}{2} = \frac{x^2 - x}{2}\)

\(6C_x^2 = 6 \cdot \frac{x^2 - x}{2} = 3 \cdot \frac{x^2 - x}{1} = 3(x^2 - x) = 3x^2 - 3x\)

\(C_x^3 = \frac{x!}{(x - 3)! \cdot 3!} = \frac{x \cdot (x - 1) \cdot (x-2) \cdot (x - 3)!}{(x - 3)! \cdot 6} = \frac{x(x - 1)(x - 2)}{6} = \frac{x(x^2 - \underline{2x} - \underline{x} + 2)}{6} = \frac{x^3 - 3x^2 + 2x}{6}\)

\(6C_x^3 = 6 \cdot \frac{x^3 - 3x^2 + 2x}{6} = x^3 - 3x^2 + 2x\)

Аз ин баробариҳо мебарояд, ки \(x \geq 3\).

\(C_x^1 = x, 6C_x^2 = 3x^2 - 3x, 6C_x^3 = x^3 - 3x^2 + 2x.\)

\(C_x^1 + 6C_x^2 + 6C_x^3 = 9x^2 - 14x \Rightarrow \underline{x} + \widetilde{3x^2} - \underline{3x} + x^3 - \widetilde{3x^2} + \underline{2x} = 9x^2 - 14x\)

\(x^3 = 9x^2 - 14x\)

\(x^3 - 9x^2 + 14x = 0\)

\(x(x^2 - 9x + 14) = 0\)

\(x_1 = 0\) ё ки \(x^2 - 9x + 14 = 0\)

\(x^2 - 9x + 14 = 0\)

\(D = (-9)^2 - 4 \cdot 1 \cdot 14 = 81 - 56 = 25 = 5^2, D>0\)

\(x_{2, 3} = \frac{-(-9) \pm \sqrt{5^2}}{2 \cdot 1} = \frac{9 \pm 5}{2}\)

\(x_2 = \frac{9 - 5}{2} = \frac{4}{2} = 2\)

\(x_3 = \frac{9 + 5}{2} = \frac{14}{2} = 7.\)

\(x_1 = 0, x_2 = 2, x_3 = 7.\)

\(0 \ngeq 3\), \(2 \ngeq 3\) ва \(7 > 3.\)

Яъне, танҳо \(x = 7\) шарти \(x \geq 3\)-ро қаноат менамояд.

Ҷавоб: x = 7.