$A=x^2–5x+6,B=x^2-7x+12,C=x^2-9x+20$ তিনটি বীজগাণিতিক রাশি।

$\frac{x}{y}$এবং $\frac{x+y}{y}$ এর বিয়োগফল: $\frac{x}{y}$-$\frac{x+y}{y}$

$=\frac{x-x-y}{y}$

$=\frac{-y}{y}$

=-1

দেওয়া আছে, B = $x^2$- 7x + 12 এবং C =$x^2$- 9x + 20

এখন, $\frac{1}{B}+\frac{1}{C}$

$=\frac{1}{x^2-7x+12}+\frac{1}{x^2-9x+20}$

$=\frac{1}{x^2-4x-3x+12}+\frac{1}{x^2-5x-4x+20}$

$=\frac{1}{x(x-4)-3(x-4)}+\frac{1}{x(x-5)-4(x-5)}$

$=\frac{1}{(x-4)(x-3)}+\frac{1}{(x-4)(x-5)}$

$=\frac{x-5+x-3}{(x-3)(x-4)(x-5)}$

$=\frac{2x-8}{(x-3)(x-4)(x-5)}$

$=\frac{2}{(x-3)(x-5)}$

দেওয়া আছে,

A =$x^2$-5x + 0.6 B = $x^2$- 0.2x +12 এবং C = $x^2$-9x +20

$\therefore$ প্রদত্ত ভগ্নাংশগুলো

$\frac{1}{A}=\frac{1}{x^2+5x+6},\frac{1}{B}=\frac{1}{x^2-7x+12}$এবং $\frac{1}{C}=\frac{1}{x^2-9x+20}$

১ম ভগ্নাংশের হর = $x^2-5x+6=x^2-3x-2x+6$

$=x(x-3)-2(x-3)=(x-3)(x-2)$

২য় ভগ্নাংশের হর=$$\begin{gathered} =x^2-7x+12=x^2-4x-3x+12 \\ =x(x-4)-3(x-4)=(x-4)(x-3) \end{gathered}$$

৩য় ভগ্নাংশের হর =$=x^2-9x+20=x^2-5x-4x+20$

$=x(x-5)-4(x-5)=(x-4)(x-5)$

$\therefore$ভগ্নাংশের হরগুলোর ল.সা.গু. =  (x-2)(x-3) (x-4)(x-5)$\frac{1}{x^2-5x+6}=\frac{1}{(x-2)(x-3)}=\frac{1\times (x-4)(x-5)}{(x-2) (x-3)\times (x-4) (x-5)}$

$=\frac{(x-4)(x-5)}{x-2) (x-3\left)\right(x-4) (x-5)}$

$\frac{1}{x^2-7x+12}=\frac{1}{(x-3)(x-4)}=\frac{1\times (x-2)(x-5)}{(x-2) (x-3)\times (x-4) (x-5)}$$=\frac{(x-2)(x-5)}{x-2) (x-3\left)\right(x-4) (x-5)}$

$\frac{1}{x^2-9x+20}=\frac{1}{(x-4)(x-5)}=\frac{1\times (x-2)(x-3)}{(x-4) (x-5)\times (x-2) (x-3)}$

$=\frac{(x-2)(x-3)}{x-2) (x-3\left)\right(x-4) (x-5)}$