Муодилаҳоро ҳал кунед:

\[C_{x+1}^{x-2}+2C_{x-1}^3=7(x-1).\]

Ҳал.

\(C_n^m = \frac{n!}{(n - m)! m!}\), ки дар ин ҷо \(m \leq n\).

\(C_n^0 = 1\).

\begin{multline}
C_{x+1}^{x-2} = \frac{(x+1)!}{(x+1-(x-2)!(x-2)!)} = \frac{(x+1)!}{(x+1-x+2)!(x-2)!}=\\
=\frac{(x+1)\cdot x\cdot (x-1)(x-2)!}{3!(x-2)!}=\frac{(x+1)\cdot x\cdot (x-1)}{6}=\\
=\frac{x\cdot(x+1)(x-1)}{6}=\frac{x(x+1)(x-1)}{6},
\end{multline}
ки дар ин ҷо \(x-2\leq x+1\).

\(C_{x-1}^3=\frac{(x-1)!}{(x-1-3)!3!}=\frac{(x-1)(x-2)(x-3)(x-4)!}{6\cdot(x-4)!}=\frac{(x-1)(x-2)(x-3)}{6}\)

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

ки дар ин ҷо \(3\leq x-1\).

Азбаски

\(C_{x+1}^{x+2}=\frac{x(x+1)(x-1)}{6}\)

ва

\(2C_{x-1}^3=\frac{(x-1)(x-2)(x-3)}{3}\),

пас

\(C_{x+1}^{x+2}+2C_{x-1}^3=\frac{x(x+1)(x-1)}{6}+\frac{(x-1)(x-2)(x-3)}{3}\)

ва

\(\frac{x(x+1)(x-1)}{6}+\frac{(x-1)(x-2)(x-3)}{3}=7(x-1)\).

Ин муодиларо ҳал менамоем:

\(\frac{x(x+1)(x-1)}{6}+\frac{(x-1)(x-2)(x-3)}{3}=7(x-1)\qquad | \cdot6\)
\(x(x+1)(x-1)+2(x-1)(x-2)(x-3)=42(x-1)\)
\((x-1)(x(x+1)+2(x-2)(x-3))=42(x-1)\qquad | \cdot\frac{1}{x-1}\)
\(x(x+1)+2(x-2)(x-3)=42\)
\(x^2+x+(2x-4)(x-3)=42\)
\(x^2+x+2x^2-6x-4x+12-42=0\)
\(x^2+2x^2+x-6x-4x+12-42=0\)
\(3x^2-9x-30=0\)
1)\(a=3;b=-9;c=-30\)
2)\( D=b^2-4\cdot a\cdot c=(-9)^2-4\cdot3\cdot(-30)=81+360=441>0\)
3)\(x_{1,2}=\frac{-b\pm\sqrt{D}}{2a}=\frac{-(-9)\pm\sqrt(441)}{2\cdot3}=\frac{9\pm21}{6}=\frac{3(3\pm7)}{6}=\frac{3\pm7}{2}\)
\(x_1=\frac{3-7}{2}=\frac{-4}{2}=-2\)
\(x_2=\frac{3+7}{2}=\frac{10}{2}=5\)

Мувофиқи шарт \(x-2\leq x+1\) ва \(3\leq x-1\).

Санҷиш.
1) \(-2-2\leq -2+1\)
\(-4<-3\)
\(3\leq -2-1\)
\(3\nleq -3\)

Қимати якуми \(x\) ба шарти додашуда мувофиқ нест, яъне \(x\neq-2\).

2) \(5-2\leq 5+1\)
\(3<6\)
\(3\leq 5-1\)
\(3<4\)

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

Ҷавоб: \(x=5\).