Муодилаҳоро ҳал кунед:

\[\frac{A_x^4}{A_{x+1}^3-C_x^{x-4}}=\frac{24}{23}.\]
Ҳал.

\(A_n^m = \frac{n!}{(n - m)!}\), ки дар ин ҷо \(m \leq n\)
\(C_n^m = \frac{n!}{(n - m)! m!}\), ки дар ин ҷо \(m \leq n\).
\(C_n^0 = 1\).
\begin{multline}
A_x^4=\frac{x!}{(x-4)!}=\frac{x(x-1)(x-2)(x-3)(x-4)!}{(x-4)!}=
\\=x(x-1)(x-2)(x-3),
\end{multline}
ки дар ин ҷо \(4\leq x\).
\begin{multline}
A_{x+1}^3=\frac{(x+1)!}{(x+1-3)!}=\frac{(x+1)\cdot x\cdot (x-1)(x-2)!}{(x-2)!}=
\\=(x+1)\cdot x\cdot(x-1)=x(x-1)(x+1)),
\end{multline}
ки дар ин ҷо \(3\leq x+1\).
\begin{multline}C_x^{x-4}=\frac{x!}{(x-(x-4))!(x-4)!}=\frac{x(x-1)(x-2)(x-3)(x-4)!}{(x-x+4)!(x-4!)}=\frac{x(x-1)(x-2)(x-3)}{4!}= \\ = \frac{x(x-1)(x-2)(x-3)}{24}=\frac{1}{24}\cdot x(x-1)(x-2)(x-3),
\end{multline}
ки дар ин ҷо \(x-4\leq x\).

Азбаски
\(A_x^4=x(x-1)(x-2)(x-3)\),
\(A_{x+1}^3=x(x-1)(x+1)\)
ва
\(C_x^{x-4}=\frac{1}{24}\cdot x(x-1)(x-2)(x-3)\),
пас
\(\frac{A_x^4}{A_{x+1}^3-C_x^{x-4}}=\frac{x(x-1)(x-2)(x-3)}{x(x-1)(x+1)-\frac{1}{24}\cdot x(x-1)(x-2)(x-3)}\)
ва
\(\frac{x(x-1)(x-2)(x-3)}{x(x-1)(x+1)-\frac{1}{24}\cdot x(x-1)(x-2)(x-3)}=\frac{24}{23}\).

Ин муодиларо ҳал менамоем:

\(\frac{x(x-1)(x-2)(x-3)}{x(x-1)(x+1)-\frac{1}{24}\cdot x(x-1)(x-2)(x-3)}=\frac{24}{23}\)

\(\frac{x(x-1)(x-2)(x-3)}{x(x-1)(x+1-\frac{1}{24}\cdot(x-2)(x-3))}=\frac{24}{23}\)

\(\frac{(x-2)(x-3)}{x+1-\frac{1}{24}\cdot(x-2)(x-3)}=\frac{24}{23}\)

\(23\cdot(x-2)(x-3)=24\cdot(x+1-\frac{1}{24}\cdot(x-2)(x-3))\)
\(23(x^2-3x-2x+6)=24x+24-(x^2-3x-2x+6)\)
\(23(x^2-5x+6)=24x+24-(x^2-5x+6)\)
\(23x^2-115x+138=24x+24-x^2+5x-6\)
\(23x^2-115x+138=29x-x^2+18\)
\(23x^2-115x+138-29x+x^2-18=0\)
\(23x^2+x^2-115x-29x+138-18=0\)
\(24x^2-144x+120=0\qquad |\cdot\frac{1}{24}\)
\(x^2-6x+5=0\)
\[1) a=1;b=-6;c=5.\]
\[2) D=b^2-4\cdot a\cdot c=(-6)^2-4\cdot 1\cdot 5=36-20=16 >0\]
\[3) x_{1,2}=\frac{-b\pm\sqrt{D}}{2a}=\frac{-(-6)\pm\sqrt{16}}{2\cdot 1}=\frac{6\pm4}{2}=\frac{2(3\pm2)}{2}=3\pm2\]
\[x_1=3-2=1; \qquad x_2=3+2=5.\]
Мувофиқи шарт \(4\leq x\), \(3\leq x+1\) ва \(x-4\leq x\).

Санҷиш.
1) \(4\nleq1\)
\(3\nleq1+1\), азбаски \(3\nleq2\).
\(1-4\leq1\)
\(-3<1\)

Қимати якуми \(x\) ба шарти додашуда мувофиқ нест, яъне \(x\neq1\).

2) \(4<5\)
\(3<5+1\), азбаски \(3<6\).
\(5-4<5\), азбаски \(1<5\).

Пас, ба шарти додашуда фақат қимати дуюми \(x\) мувофиқ аст, яъне \(x=5\).

Ҷавоб: \(x=5\).