Страница 329 номер 1091, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(C_{x + 1}^{2} + C_{x + 1}^{3} = 7x\)
\(x + 1 \geqslant 3, x \geqslant 2;\)
\(C_{x + 2}^{3} = 7x,\)
\(\dfrac{(x + 2)!}{3!(x + 2 - 3)!} = 7x,\)
\(\dfrac{(x + 2)!}{3!(x - 1)!} = 7x,\)
\(x(x + 1)(x + 2) = 7x \cdot 6,\)
\(x^{2} + 3x - 40 = 0,\)
\(x_{1{,}2} = \dfrac{-3 \pm \sqrt{9 + 160}}{2},\)
\(x_{1} = 5\), \(x_{2} = - 8\) – не подходит;

Ответ: \(x = 5\)

2) \(C_{x - 1}^{3} + C_{x - 1}^{2} = 4(x - 1)\)
\(x - 1 \geqslant 3, x \geqslant 4;\)
\(C_{x}^{3} = 4(x - 1),\)
\(\dfrac{x!}{3! \cdot (x - 3)!} = 4(x - 1),\)
\(x(x - 1)(x - 2) = 4(x - 1) \cdot 6,\)
\(x^{2} - 2x - 24 = 0,\)
\(x_{1{,}2} = \dfrac{2 \pm \sqrt{4 + 96}}{2},\)
\(x_{1} = 6\), \(x_{2} = - 4\) – не подходит;

Ответ: \(x = 6\)

3) \(C_{x}^{3} = \frac{4}{15}C_{x + 2}^{4}\)
\(x \geqslant 3;\)
\(\dfrac{x!}{3!(x - 3)!} = \dfrac{4}{15} \cdot \dfrac{(x + 2)!}{4!(x + 2 - 4)!},\)
\(x(x - 1)(x - 2) = \dfrac{4 \cdot 3!}{15} \cdot \dfrac{x(x - 1)(x + 1)(x + 2)}{4!},\)
\(15(x - 2) = (x + 1)(x + 2),\)
\(15x - 30 = x^{2} + 3x + 2,\)
\(x^{2} - 12x + 32 = 0,\)
\(x_{1{,}2} = \dfrac{12 \pm \sqrt{144 - 128}}{2},\)
\(x_{1} = 8, x_{2} = 4;\)

4) \(5C_{x}^{3} = C_{x + 2}^{4}\)
\(x \geqslant 3;\)
\(5\dfrac{x!}{3!(x - 3)!} = \dfrac{(x + 2)!}{4!(x - 2)!},\)
\(5x(x - 1)(x - 2) = \dfrac{x(x - 1)(x + 1)(x + 2)}{4},\)
\(20(x - 2) = (x + 1)(x + 2),\)
\(20x - 40 = x^{2} + 3x + 2,\)
\(x^{2} - 17x + 42 = 0,\)
\(x_{1{,}2} = \dfrac{17 \pm \sqrt{289 - 168}}{2},\)
\(x_{1} = 14, x_{2} = 3;\)

5) \(C_{3x + 1}^{3x - 1} = 120\)
\(3x - 1 \geqslant 0, x \geqslant \dfrac{1}{3};\)
\(\dfrac{(3x + 1)!}{(3x - 1)!(3x + 1 - 3x + 1)!} = 120,\)
\(\dfrac{3x(3x + 1)}{2!} = 120,\)
\(9x^{2} + 3x = 240,\)
\(3x^{2} + x - 80 = 0,\)
\(x_{1{,}2} = \dfrac{-1 \pm \sqrt{1 + 960}}{6},\)
\(x_{1} = 5\), \(x_{2} = - \frac{32}{6}\) – не подходит;

Ответ: \(x = 5\)

6) \(C_{2x + 1}^{2x - 1} = 36\)
\(2x - 1 \geqslant 0, x \geqslant \dfrac{1}{2};\)
\(\dfrac{(2x + 1)!}{(2x - 1)!(2x + 1 - 2x + 1)!} = 36,\)
\(\dfrac{2x(2x + 1)}{2!} = 36,\)
\(4x^{2} + 2x = 72,\)
\(2x^{2} + x - 36 = 0,\)
\(x_{1{,}2} = \dfrac{-1 \pm \sqrt{1 + 288}}{4},\)
\(x_{1} = 4\), \(x_{2} = - \frac{18}{4}\) – не подходит;

Ответ: \(x = 4\).