Hoán vị - Chỉnh hợp - Tổ hợp - Giải phương trình

+ Số đoạn thẳng tạo bởi $n$ đỉnh là $C_n^2$, trong đó có $n$ cạnh, suy ra số đường chéo là $C_n^2 - n$.

+ Đa giác đã cho có $135$ đường chéo nên $C_n^2 - n = 135$.

+ Giải PT :$\dfrac{{n!}}{{\left( {n - 2} \right)!2!}} - n = 135\,\,,\,\,\left( {n \in \mathbb{N},n \ge 2} \right)$$\Leftrightarrow \left( {n - 1} \right)n - 2n = 270$$\Leftrightarrow {n^2} - 3n - 270 = 0$$\Leftrightarrow \left[ \begin{array}{l}n = 18\,\left( TM \right)\\n =  - 15\,\left( L \right)\end{array} \right.$$\Leftrightarrow n = 18$.

ĐK: $n \in N;n \ge 4$

$\begin{array}{l}\dfrac{{A_n^4}}{{A_{n + 1}^3 - C_n^{n - 4}}} = \dfrac{{24}}{{23}} \Leftrightarrow \dfrac{{A_n^4}}{{A_{n + 1}^3 - C_n^4}} = \dfrac{{24}}{{23}}\\ \Leftrightarrow \dfrac{{\dfrac{{n!}}{{\left( {n - 4} \right)!}}}}{{\dfrac{{\left( {n + 1} \right)!}}{{\left( {n - 2} \right)!}} - \dfrac{{n!}}{{\left( {n - 4} \right)!4!}}}} = \dfrac{{24}}{{23}}\\ \Leftrightarrow \dfrac{{\dfrac{{n!}}{{\left( {n - 4} \right)!}}}}{{\dfrac{{n!\left( {n + 1} \right)}}{{\left( {n - 4} \right)!\left( {n - 3} \right)\left( {n - 2} \right)}} - \dfrac{{n!}}{{\left( {n - 4} \right)!4!}}}} = \dfrac{{24}}{{23}}\\ \Leftrightarrow \dfrac{{\dfrac{{n!}}{{\left( {n - 4} \right)!}}}}{{\dfrac{{n!}}{{\left( {n - 4} \right)!}}\left( {\dfrac{{n + 1}}{{\left( {n - 3} \right)\left( {n - 2} \right)}} - \dfrac{1}{{24}}} \right)}} = \dfrac{{24}}{{23}}\\ \Leftrightarrow \dfrac{1}{{\dfrac{{n + 1}}{{\left( {n - 3} \right)\left( {n - 2} \right)}} - \dfrac{1}{{24}}}} = \dfrac{{24}}{{23}}\\ \Leftrightarrow \dfrac{{24\left( {n - 3} \right)\left( {n - 2} \right)}}{{24n + 24 - \left( {n - 3} \right)\left( {n - 2} \right)}} = \dfrac{{24}}{{23}}\\ \Leftrightarrow 23\left( {{n^2} - 5n + 6} \right) = 24n + 24 - \left( {{n^2} - 5n + 6} \right)\\ \Leftrightarrow 24{n^2} - 144n + 120 = 0 \Leftrightarrow \left[ \begin{array}{l}n = 5\,\,\left( {tm} \right)\\n = 1\,\,\,\,\left( {ktm} \right)\end{array} \right.\end{array}$

Với $n = 5$ ta có: $P = {\left( {5 + 1} \right)^2} - 3.5 + 5 = 26$

$\begin{array}{l}12C_x^1 + C_{x + 4}^2 = 162\\ \Leftrightarrow 12x + \dfrac{{\left( {x + 4} \right)!}}{{2!\left( {x + 2} \right)!}} = 162\\ \Leftrightarrow 12x + \dfrac{{\left( {x + 4} \right)\left( {x + 3} \right)}}{2} = 162\\ \Leftrightarrow 24x + {x^2} + 7x + 12 = 324\\ \Leftrightarrow {x^2} + 31x - 312 = 0 \Leftrightarrow \left[ \begin{array}{l}x = 8\,\,\,\,\,\,\,\,\,\left( {tm} \right)\\x =  - 39\,\,\left( {ktm} \right)\end{array} \right.\end{array}$

$\Rightarrow A_{x - 1}^2 - C_x^1 = A_7^2 - C_8^1 = 34$

ĐK: $x \ge 3,x \in N$

$\begin{array}{l}C_x^1 + C_x^2 + C_x^3 = \dfrac{7}{2}x\\ \Leftrightarrow x + \dfrac{{x!}}{{2!\left( {x - 2} \right)!}} + \dfrac{{x!}}{{3!\left( {x - 3} \right)!}} = \dfrac{7}{2}x\\ \Leftrightarrow x + \dfrac{{x\left( {x - 1} \right)}}{2} + \dfrac{{x\left( {x - 1} \right)\left( {x - 2} \right)}}{6} = \dfrac{7}{2}x\\ \Leftrightarrow 6x + 3{x^2} - 3x + {x^3} - 3{x^2} + 2x = 21x\\ \Leftrightarrow {x^3} - 16x = 0\\ \Leftrightarrow x\left( {{x^2} - 16} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}x = 0\,\,\,\,\,\,\,\left( {ktm} \right)\\x =  - 4\,\,\,\,\left( {ktm} \right)\\x = 4\,\,\,\,\,\,\,\left( {tm} \right)\end{array} \right.\end{array}$

ĐK: $\left\{ \begin{array}{l}x \ge 1\\x + 1 \ge 2\\x + 4 \ge 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge 1\\x \ge  - 3\end{array} \right. \Leftrightarrow x \ge 1,x \in N$

$\begin{array}{l}\dfrac{1}{{C_x^1}} - \dfrac{1}{{C_{x + 1}^2}} = \dfrac{7}{{6C_{x + 4}^1}}\\ \Leftrightarrow \dfrac{1}{x} - \dfrac{1}{{\dfrac{{\left( {x + 1} \right)!}}{{2!\left( {x - 1} \right)!}}}} = \dfrac{7}{{6\left( {x + 4} \right)}}\\ \Leftrightarrow \dfrac{1}{x} - \dfrac{2}{{x\left( {x + 1} \right)}} = \dfrac{7}{{6\left( {x + 4} \right)}}\\ \Leftrightarrow \dfrac{{6\left( {x + 1} \right)\left( {x + 4} \right) - 12\left( {x + 4} \right) - 7x\left( {x + 1} \right)}}{{6x\left( {x + 1} \right)\left( {x + 4} \right)}} = 0\\ \Leftrightarrow 6{x^2} + 30x + 24 - 12x - 48 - 7{x^2} - 7x = 0\\ \Leftrightarrow \left[ \begin{array}{l}x = 8\,\,\left( {tm} \right)\\x = 3\,\,\left( {tm} \right)\end{array} \right.\end{array}$

Vậy có $2$ giá trị của $x$ thỏa mãn bài toán.

ĐK: $\left\{ \begin{array}{l}2x \ge 2\\x \ge 2\\x \ge 3\end{array} \right. \Leftrightarrow x \ge 3,x \in N$

$\begin{array}{l}\dfrac{1}{2}A_{2x}^2 - A_x^2 \le \dfrac{6}{x}C_x^3 + 10\\ \Leftrightarrow \dfrac{1}{2}\dfrac{{\left( {2x} \right)!}}{{\left( {2x - 2} \right)!}} - \dfrac{{x!}}{{\left( {x - 2} \right)!}} \le \dfrac{6}{x}\dfrac{{x!}}{{3!\left( {x - 3} \right)!}} + 10\\ \Leftrightarrow \dfrac{{\left( {2x - 1} \right)2x}}{2} - x\left( {x - 1} \right) \le \left( {x - 1} \right)\left( {x - 2} \right) + 10\\ \Leftrightarrow 2{x^2} - x - {x^2} + x - {x^2} + 3x - 2 - 10 \le 0\\ \Leftrightarrow 3x - 12 \le 0\\ \Leftrightarrow x \le 4\end{array}$

 Kết hợp điều kiện ta có $3 \le x \le 4$

Mà $x \in Z \Rightarrow \left[ \begin{array}{l}{x_1} = 3\\{x_2} = 4\end{array} \right. \Rightarrow {x_1}.{x_2} = 3.4 = 12$ 

ĐK: $\left\{ \begin{array}{l}x \ge y \ge 0\\x \ge y + 1 \ge 0\\x \ge y - 1 \ge 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}y \ge 1\\x \ge 2\end{array} \right.\,\,\left( {x,y \in N} \right)$

$\begin{array}{l}\left\{ \begin{array}{l}C_x^y - C_x^{y + 1} = 0\,\,\,\,\,\,\,\,\,\left( 1 \right)\\4C_x^y - 5C_x^{y - 1} = 0\,\,\,\,\left( 2 \right)\end{array} \right.\\\left( 1 \right) \Leftrightarrow C_x^y = C_x^{y + 1} \Leftrightarrow \left\{ \begin{array}{l}y + 1 = y\\y + 1 = x - y\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}0 = 1\,\,\left( {VN} \right)\\x = 2y + 1\end{array} \right. \Rightarrow x = 2y + 1\\\left( 2 \right) \Leftrightarrow 4C_x^y = 5C_x^{y - 1}\\ \Leftrightarrow \dfrac{{4x!}}{{y!\left( {x - y} \right)!}} = \dfrac{{5x!}}{{\left( {y - 1} \right)!\left( {x - y + 1} \right)!}}\\ \Leftrightarrow \dfrac{4}{y} = \dfrac{5}{{x - y + 1}}\,\left( * \right)\end{array}$

Thay x = 2y + 1 vào phương trình (*) ta được: $\dfrac{4}{y} = \dfrac{5}{{2y + 1 - y + 1}} \Leftrightarrow 4\left( {y + 2} \right) = 5y \Leftrightarrow y = 8\,\,\left( {tm} \right)$

$\Rightarrow x = 2.8 + 1 = 17\,\,\left( {tm} \right)$

Vậy hệ phương trình có 1 nghiệm là $\left( {x;y} \right) = \left( {17;8} \right)$  

ĐK: $\left\{ \begin{array}{l}x + 1 \ge 2\\x \ge 2\end{array} \right. \Leftrightarrow x \ge 2\,\,\left( {x \in N} \right)$

$\begin{array}{l}2C_{x + 1}^2 + 3A_x^2 < 30\\ \Leftrightarrow \dfrac{{2\left( {x + 1} \right)!}}{{2!\left( {x - 1} \right)!}} + \dfrac{{3x!}}{{\left( {x - 2} \right)!}} < 30\\ \Leftrightarrow \left( {x + 1} \right)x + 3x\left( {x - 1} \right) < 30\\ \Leftrightarrow {x^2} + x + 3{x^2} - 3x < 30\\ \Leftrightarrow 4{x^2} - 2x - 30 < 0\\ \Leftrightarrow x \in \left( { - \dfrac{5}{2};3} \right)\end{array}$

Mà $x \ge 2,x \in \mathbb{N}$ nên bất phương trình chỉ có nghiệm $x = 2$

Đáp án A: $\dfrac{{x - 2}}{{x - 3}} \le 0 \Leftrightarrow 2 \le x < 3$

Mà $x \in \mathbb{N}$ nên $x = 2$

Đáp án B: ${x^2} - 5x + 6 < 0$ $\Leftrightarrow \left( {x - 2} \right)\left( {x - 3} \right) < 0 \Leftrightarrow 2 < x < 3$

Do $x \in \mathbb{N}$ nên không có giá trị nào của $x$ thỏa mãn bất phương trình.

Đáp án C: $\dfrac{{{x^2} - 4}}{{x - 3}} \le 0$ $\Leftrightarrow \dfrac{{\left( {x - 2} \right)\left( {x + 2} \right)}}{{x - 3}} \le 0$ $\Leftrightarrow \left[ \begin{array}{l}x \le  - 2\\2 \le x < 3\end{array} \right.$

Do $x \in \mathbb{N}$ nên $x = 2$

Đáp án D: $\dfrac{{\left( {x - 2} \right)\left( {{x^2} + 1} \right)}}{{x - 3}} \le 0$ $\Leftrightarrow \dfrac{{x - 2}}{{x - 3}} \le 0$ $\Leftrightarrow 2 \le x < 3$

Do $x \in \mathbb{N}$ nên $x = 2$

Vậy các đáp án A, C, D đều có tập nghiệm $S = \left\{ 2 \right\}$ trùng với tập nghiệm của bất phương trình bài cho.

Chi có đáp án B là tập nghiệm $\emptyset$

ĐK: $n \ge 3,n \in N$

$\begin{array}{l}A_n^3 + 5A_n^2 = 2\left( {n + 15} \right)\\ \Leftrightarrow \dfrac{{n!}}{{\left( {n - 3} \right)!}} + 5\dfrac{{n!}}{{\left( {n - 2} \right)!}} = 2\left( {n + 15} \right)\\ \Leftrightarrow n\left( {n - 1} \right)\left( {n - 2} \right) + 5n\left( {n - 1} \right) = 2\left( {n + 15} \right)\\ \Leftrightarrow {n^3} - 3{n^2} + 2n + 5{n^2} - 5n - 2n - 30 = 0\\ \Leftrightarrow {n^3} + 2{n^2} - 5n - 30 = 0\\ \Leftrightarrow n = 3\,\,\left( {tm} \right)\end{array}$

ĐK: $x \ge 3,x \in N$.

Phương trình đã cho có dạng

$\begin{array}{l}\dfrac{{x!}}{{\left( {x - 3} \right)!}} + \dfrac{{2\left( {x + 1} \right)!}}{{2!\left( {x - 1} \right)!}} - \dfrac{{3\left( {x - 1} \right)!}}{{2!\left( {x - 3} \right)!}} = 3{x^2} + 6! + 159\\ \Leftrightarrow x\left( {x - 1} \right)\left( {x - 2} \right) + x\left( {x + 1} \right) - \dfrac{3}{2}\left( {x - 1} \right)\left( {x - 2} \right) = 3{x^2} + 879\\ \Leftrightarrow x = 12\,\,\left( {tm} \right)\end{array}$

(Dùng lệnh SHIFT SLOVE trên máy tính)

ĐK: $0 \le k \le 12\,\,\left( {k \in N} \right)$

$\begin{array}{l}C_{14}^k + C_{14}^{k + 2} = 2C_{14}^{k + 1}\\ \Leftrightarrow \frac{{14!}}{{k!\left( {14 - k} \right)!}} + \frac{{14!}}{{\left( {k + 2} \right)!\left( {12 - k} \right)!}} = 2\frac{{14!}}{{\left( {k + 1} \right)!\left( {13 - k} \right)!}}\\ \Leftrightarrow \frac{{14!}}{{k!\left( {12 - k} \right)!}}\left[ {\frac{1}{{\left( {14 - k} \right)\left( {13 - k} \right)}} + \frac{1}{{\left( {k + 2} \right)\left( {k + 1} \right)}} - \frac{2}{{\left( {k + 1} \right)\left( {13 - k} \right)}}} \right] = 0\\ \Leftrightarrow \frac{1}{{\left( {14 - k} \right)\left( {13 - k} \right)}} + \frac{1}{{\left( {k + 2} \right)\left( {k + 1} \right)}} - \frac{2}{{\left( {k + 1} \right)\left( {13 - k} \right)}} = 0\\ \Leftrightarrow \left( {k + 2} \right)\left( {k + 1} \right) + \left( {14 - k} \right)\left( {13 - k} \right) - 2\left( {k + 2} \right)\left( {14 - k} \right) = 0\\ \Leftrightarrow {k^2} + 3k + 2 + {k^2} - 27k + 182 + 2{k^2} - 24k - 56 = 0\\ \Leftrightarrow 4{k^2} - 48k + 128 = 0\\ \Leftrightarrow \left[ \begin{array}{l}k = 8\,\,\left( {tm} \right)\\k = 4\,\,\left( {tm} \right)\end{array} \right.\end{array}$

Vậy có $2$ giá trị của $k$ thỏa mãn yêu cầu đề bài.

ĐK: $x \ge y \ge 0,x,y \in N$

Đặt $a = A_x^y\,\,;\,\,y = C_x^y$ ta được $\left\{ \begin{array}{l}2x + 5y = 90\\5x - 2y = 80\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 20\\b = 10\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}A_x^y = 20\\C_x^y = 10\end{array} \right.$

Ta có: $C_x^y = \dfrac{{A_x^y}}{{y!}} \Leftrightarrow 10 = \dfrac{{20}}{{y!}} \Leftrightarrow y! = 2 \Leftrightarrow y = 2$

$\begin{array}{l} \Rightarrow C_x^2 = 20 \Leftrightarrow \dfrac{{x!}}{{\left( {x - 2} \right)!}} = 20 \Leftrightarrow x\left( {x - 1} \right) = 20\\ \Leftrightarrow {x^2} - x - 20 = 0 \Leftrightarrow \left[ \begin{array}{l}x = 5\,\,\,\,\,\,\left( {tm} \right)\\x =  - 4\,\,\left( {ktm} \right)\end{array} \right.\\ \Rightarrow xy = 5.2 = 10\end{array}$

ĐK: $\left\{ \begin{array}{l}0 \le x \le y\\0 \le x \le y + 2\end{array} \right. \Leftrightarrow 0 \le x \le y\,\,\left( {x,y \in N} \right)$

$\begin{array}{l}\left\{ \begin{array}{l}C_y^x:C_{y + 2}^x = \dfrac{1}{3}\\C_y^x:A_y^x = \dfrac{1}{{24}}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\dfrac{{y!}}{{x!\left( {y - x} \right)!}}.\dfrac{{x!\left( {y + 2 - x} \right)!}}{{\left( {y + 2} \right)!}} = \dfrac{1}{3}\\\dfrac{{y!}}{{x!\left( {y - x} \right)!}}.\dfrac{{\left( {y - x} \right)!}}{{y!}} = \dfrac{1}{{24}}\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}\dfrac{{\left( {y - x + 1} \right)\left( {y - x + 2} \right)}}{{\left( {y + 1} \right)\left( {y + 2} \right)}} = \dfrac{1}{3}\\\dfrac{1}{{x!}} = \dfrac{1}{{24}}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 4\,\,\left( {tm} \right)\\\dfrac{{\left( {y - 3} \right)\left( {y - 2} \right)}}{{\left( {y + 1} \right)\left( {y + 2} \right)}} = \dfrac{1}{3}\,\,\left( * \right)\end{array} \right.\\\left( * \right) \Leftrightarrow 3{y^2} - 15y + 18 = {y^2} + 3y + 2\\ \Leftrightarrow 2{y^2} - 18y + 16 = 0 \Leftrightarrow \left[ \begin{array}{l}y = 8\,\,\left( {tm} \right)\\y = 1\,\,\,\left( {ktm} \right)\end{array} \right.\end{array}$

Vậy hệ phương trình có 1 nghiệm $\left( {x;y} \right) = \left( {4;8} \right)$

$A_{n + k}^{n + 1} + A_{n + k}^{n + 2} = \dfrac{{\left( {n + k} \right)!}}{{\left( {k - 1} \right)!}} + \dfrac{{\left( {n + k} \right)!}}{{\left( {k - 2} \right)!}}$ $= \dfrac{{\left( {n + k} \right)!\left( {1 + k - 1} \right)}}{{\left( {k - 1} \right)!}}$ $= k.\dfrac{{\left( {n + k} \right)!}}{{\left( {k - 1} \right)!}}$ $= {k^2}\dfrac{{\left( {n + k} \right)!}}{{k!}} = {k^2}A_{n + k}^n$

ĐK: $\left\{ \begin{array}{l}n - 1 \ge 4\\n - 1 \ge 3\\n - 2 \ge 2\end{array} \right. \Leftrightarrow n \ge 5,n \in N$

$\begin{array}{l}C_{n - 1}^4 - C_{n - 1}^3 - \dfrac{5}{4}A_{n - 2}^2 < 0\\ \Leftrightarrow \dfrac{{\left( {n - 1} \right)!}}{{4!\left( {n - 5} \right)!}} - \dfrac{{\left( {n - 1} \right)!}}{{3!\left( {n - 4} \right)!}} - \dfrac{5}{4}\dfrac{{\left( {n - 2} \right)!}}{{\left( {n - 4} \right)!}} < 0\\ \Leftrightarrow \dfrac{{\left( {n - 2} \right)!}}{{\left( {n - 5} \right)!}}\left( {\dfrac{{n - 1}}{{4!}} - \dfrac{{n - 1}}{{3!\left( {n - 4} \right)}} - \dfrac{5}{{4\left( {n - 4} \right)}}} \right) < 0\\ \Leftrightarrow \dfrac{{n - 1}}{{24}} - \dfrac{{n - 1}}{{6\left( {n - 4} \right)}} - \dfrac{5}{{4\left( {n - 4} \right)}} < 0\\ \Leftrightarrow \dfrac{{\left( {n - 1} \right)\left( {n - 4} \right) - 4\left( {n - 1} \right) - 30}}{{24\left( {n - 4} \right)}} < 0\end{array}$

Vì $n \ge 5 \Rightarrow n - 4 > 0$ nên

$bpt \Leftrightarrow \left\{ \begin{array}{l}\left( {n - 1} \right)\left( {n - 4} \right) - 4\left( {n - 1} \right) - 30 < 0\\n \ge 5\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{n^2} - 9n - 22 < 0\\n \ge 5\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} - 2 < n < 11\\n \ge 5\end{array} \right. \Leftrightarrow 5 \le n < 11$

Vì $n \in N \Rightarrow n \in \left\{ {5;6;7;8;9;10} \right\}$

Vậy có 6 giá trị của n thỏa mãn yêu cầu bài toán.

ĐK: $x \ge 2,y \ge 3,x,y \in N$

Ta có: $C_x^2 = \dfrac{1}{{2!}}A_x^2 = \dfrac{1}{2}A_x^2\,;\,C_y^3 = \dfrac{1}{{3!}}A_y^3 = \dfrac{1}{6}A_y^3$

Đặt $A_x^2 = a\,;\,A_y^3 = b$ ta có:

$hpt \Leftrightarrow \left\{ \begin{array}{l}a + \dfrac{b}{6} = 22\\b + \dfrac{a}{2} = 66\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 12\\b = 60\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}A_x^2 = 12\,\,\left( 1 \right)\\A_y^3 = 60\,\,\left( 2 \right)\end{array} \right.$

Giải (1):

$A_x^2 = 12 \Leftrightarrow \dfrac{{x!}}{{\left( {x - 2} \right)!}} = 12 \Leftrightarrow x\left( {x - 1} \right) = 12 \Leftrightarrow {x^2} - x - 12 = 0 \Leftrightarrow \left[ \begin{array}{l}x = 4\,\,\,\,\,\left( {tm} \right)\\x =  - 3\,\,\left( {ktm} \right)\end{array} \right.$

Giải (2): 

$\begin{array}{l}A_y^3 = 60 \Leftrightarrow \dfrac{{y!}}{{\left( {y - 3} \right)!}} = 60\\ \Leftrightarrow y\left( {y - 1} \right)\left( {y - 2} \right) = 60\\ \Leftrightarrow {y^3} - 3{y^2} + 2y - 60 = 0\\ \Leftrightarrow y = 5\,\,\left( {tm} \right)\end{array}$

Vậy $x - y = 4 - 5 =  - 1$ 

ĐK: $\left\{ \begin{array}{l}x + 1 \ge y \ge 0\\x \ge y + 1 \ge 0\\x \ge y - 1 \ge 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}y \ge 1\\x \ge y + 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}y \ge 1\\x \ge 2\end{array} \right.\,\,\left( {x,y \in N} \right)$

$\begin{array}{l}C_{x + 1}^y:C_x^{y + 1}:C_x^{y - 1} = 6:5:2 \Rightarrow \left\{ \begin{array}{l}\dfrac{{C_{x + 1}^y}}{{C_x^{y + 1}}} = \dfrac{6}{5}\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\\\dfrac{{C_{x + 1}^y}}{{C_x^{y - 1}}} = \dfrac{6}{2} = 3\,\,\left( 2 \right)\end{array} \right.\\\left( 1 \right) \Leftrightarrow \dfrac{{\dfrac{{\left( {x + 1} \right)!}}{{y!\left( {x + 1 - y} \right)!}}}}{{\dfrac{{x!}}{{\left( {y + 1} \right)!\left( {x - y - 1} \right)!}}}} = \dfrac{6}{5}\\ \Leftrightarrow \dfrac{{\left( {x + 1} \right)!}}{{y!\left( {x + 1 - y} \right)!}}.\dfrac{{\left( {y + 1} \right)!\left( {x - y - 1} \right)!}}{{x!}} = \dfrac{6}{5}\\ \Leftrightarrow \dfrac{{\left( {x + 1} \right)\left( {y + 1} \right)}}{{\left( {x - y} \right)\left( {x - y + 1} \right)}} = \dfrac{6}{5}\,\,\,\left( 3 \right)\\\left( 2 \right) \Leftrightarrow \dfrac{{\dfrac{{\left( {x + 1} \right)!}}{{y!\left( {x + 1 - y} \right)!}}}}{{\dfrac{{x!}}{{\left( {y - 1} \right)!\left( {x - y + 1} \right)!}}}} = 3\\ \Leftrightarrow \dfrac{{\left( {x + 1} \right)!}}{{y!\left( {x + 1 - y} \right)!}}\dfrac{{\left( {y - 1} \right)!\left( {x - y + 1} \right)!}}{{x!}} = 3\\ \Leftrightarrow \dfrac{{x + 1}}{y} = 3 \Rightarrow x = 3y - 1\end{array}$

Thay vào (3) ta có:

$\begin{array}{l}\dfrac{{3y\left( {y + 1} \right)}}{{\left( {2y - 1} \right)2y}} = \dfrac{6}{5}\\ \Leftrightarrow \dfrac{{y + 1}}{{4y - 2}} = \dfrac{2}{5} \Leftrightarrow 5y + 5 = 8y - 4\\ \Leftrightarrow 3y = 9 \Leftrightarrow y = 3\,\,\left( {tm} \right) \Rightarrow x = 8\,\,\left( {tm} \right)\\ \Rightarrow x + y = 11\end{array}$

Trước hết ta chứng minh công thức $C_n^k + C_n^{k + 1} = C_{n + 1}^{k + 1}$

$\begin{array}{l}VT = C_n^k + C_n^{k + 1}\\ = \dfrac{{n!}}{{k!\left( {n - k} \right)!}} + \dfrac{{n!}}{{\left( {k + 1} \right)!\left( {n - k - 1} \right)!}}\\ = \dfrac{{n!}}{{k!\left( {n - k - 1} \right)!}}\left( {\dfrac{1}{{n - k}} + \dfrac{1}{{k + 1}}} \right)\\ = \dfrac{{n!}}{{k!\left( {n - k - 1} \right)!}}.\dfrac{{k + 1 + n - k}}{{\left( {n - k} \right)\left( {k + 1} \right)}}\\ = \dfrac{{n!\left( {n + 1} \right)}}{{k!\left( {k + 1} \right)\left( {n - k - 1} \right)!\left( {n - k} \right)}}\\ = \dfrac{{\left( {n + 1} \right)!}}{{\left( {k + 1} \right)!\left( {n - k} \right)!}} = C_{n + 1}^{k + 1} = VP\end{array}$

Ta tính giá trị của biểu thức B sau đây:

$\begin{array}{l}B = C_n^k + 4C_n^{k - 1} + 6C_n^{k - 2} + 4C_n^{k - 3} + C_n^{k - 4}\\\,\,\,\,\, = C_n^k + C_n^{k - 1} + 3\left( {C_n^{k - 1} + C_n^{k - 2}} \right) + 3\left( {C_n^{k - 2} + C_n^{k - 3}} \right) + C_n^{k - 3} + C_n^{k - 4}\\\,\,\,\,\, = C_{n + 1}^k + 3C_{n + 1}^{k - 1} + 3C_{n + 1}^{k - 2} + C_{n + 1}^{k - 3}\\\,\,\,\,\, = C_{n + 1}^k + C_{n + 1}^{k - 1} + 2\left( {C_{n + 1}^{k - 1} + C_{n + 1}^{k - 2}} \right) + C_{n + 1}^{k - 2} + C_{n + 1}^{k - 3}\\\,\,\,\,\, = C_{n + 2}^k + 2C_{n + 2}^{k - 1} + C_{n + 2}^{k - 2}\\\,\,\,\,\, = C_{n + 1}^k + C_{n + 1}^{k - 1} + C_{n + 1}^{k - 1} + C_{n + 1}^{k - 2}\\\,\,\,\,\, = C_{n + 3}^k + C_{n + 3}^{k - 1}\\\,\,\,\,\, = C_{n + 4}^k\\ \Rightarrow A = B - C_{n + 4}^k + 1 = C_{n + 4}^k - C_{n + 4}^k + 1 = 1\end{array}$

Trước hết ta chứng minh $C_n^k + C_n^{k + 1} = C_{n + 1}^{k + 1}$

$\begin{array}{l}VT = C_n^k + C_n^{k + 1}\\ = \dfrac{{n!}}{{k!\left( {n - k} \right)!}} + \dfrac{{n!}}{{\left( {k + 1} \right)!\left( {n - k - 1} \right)!}}\\ = \dfrac{{n!}}{{k!\left( {n - k - 1} \right)!}}\left( {\dfrac{1}{{n - k}} + \dfrac{1}{{k + 1}}} \right)\\ = \dfrac{{n!}}{{k!\left( {n - k - 1} \right)!}}.\dfrac{{k + 1 + n - k}}{{\left( {n - k} \right)\left( {k + 1} \right)}}\\ = \dfrac{{n!\left( {n + 1} \right)}}{{k!\left( {k + 1} \right)\left( {n - k - 1} \right)!\left( {n - k} \right)}}\\ = \dfrac{{\left( {n + 1} \right)!}}{{\left( {k + 1} \right)!\left( {n - k} \right)!}} = C_{n + 1}^{k + 1} = VP\end{array}$

Ta có:

$\begin{array}{l}\,\,\,\,C_n^k + 2C_n^{k + 1} + C_n^{k + 2}\\ = C_n^k + C_n^{k + 1} + C_n^{k + 1} + C_n^{k + 2}\\ = C_{n + 1}^{k + 1} + C_{n + 1}^{k + 2}\\ = C_{n + 2}^{k + 2}\\\,\,\,\,C_n^k + 3C_n^{k + 1} + 3C_n^{k + 2} + C_n^{k + 3}\\ = C_n^k + C_n^{k + 1} + 2\left( {C_n^{k + 1} + C_n^{k + 2}} \right) + C_n^{k + 2} + C_n^{k + 3}\\ = C_{n + 1}^{k + 1} + 2C_{n + 1}^{k + 2} + C_{n + 1}^{k + 3}\\ = C_{n + 1}^{k + 1} + C_{n + 1}^{k + 2} + C_{n + 1}^{k + 2} + C_{n + 1}^{k + 3}\\ = C_{n + 2}^{k + 2} + C_{n + 2}^{k + 3}\\ = C_{n + 3}^{k + 3}\\ \Rightarrow 2C_n^k + 5C_n^{k + 1} + 4C_n^{k + 2} + C_n^{k + 3}= C_{n + 2}^{k + 2} + C_{n + 3}^{k + 3}\end{array}$

$5! - {P_4} = 5! - 4! = 96$.