Монгол Бодлогын Сан

Заавар: $$\dfrac{1}{(x-3)(x-2)(x-1)}=\dfrac{A}{x-3}+\dfrac{B}{x-2}+\dfrac{C}{x-1}$$ бол $$\dfrac{1}{(x-3)(x-2)(x-1)}=-\dfrac{A}{3}\cdot\dfrac{1}{1-\frac{x}{3}}-\dfrac{B}{2}\cdot\dfrac{1}{1-\frac{x}{2}}-C\cdot\dfrac{1}{1-x}$$ болно.

Бодолт: $$\dfrac{1}{(x-3)(x-2)(x-1)}=\dfrac{A}{x-3}+\dfrac{B}{x-2}+\dfrac{C}{x-1}$$ бол $$1=A(x-2)(x-1)+B(x-3)(x-1)+C(x-2)(x-3)$$ болно. $x=1,2,3$ утгуудыг орлуулбал $2C=1$, $-B=1$, $2A=1$ болох тул $A=\dfrac12$, $B=-1$, $C=\dfrac12$ байна. Иймд \begin{align*} \dfrac{1}{(x-3)(x-2)(x-1)}&=\dfrac{\frac12}{x-3}-\dfrac{1}{x-2}+\dfrac{\frac12}{x-1}\\ &=-\frac16\cdot\dfrac{1}{1-\frac{x}{3}}+\dfrac12\cdot\dfrac{1}{1-\frac{x}{2}}-\dfrac12\cdot\dfrac{1}{1-x} \end{align*} $$\dfrac{1}{1-\frac{x}{a}}=1+\dfrac{x}{a}+\dfrac{x^2}{a^2}+\dfrac{x^3}{a^3}+\cdots$$ тул $x^2$-ийн өмнөх коэффициент нь $$-\dfrac16\cdot\dfrac{1}{3^2}+\dfrac12\cdot\dfrac{1}{2^2}-\dfrac{1}{2}\cdot \dfrac{1}{1^2}=-\dfrac{85}{216}$$ байна.