$A=\frac{1}{x^2-5x+6}, B=\frac{2}{x^2+x-12}, C=\frac{3}{x^2+2x-8}$

প্রদত্ত ভগ্নাংশ:

∴ ভগ্নাংশটির লঘিষ্ঠ রূপ: $\frac{x}{-(4y+x)}$

এখানে, A ভগ্নাংশের হর:

$$x^2 - 5x + 6 \\ = x^2 - 2x - 3x + 6 \\ = x(x - 2) - 3(x - 2) \\ = (x - 2)(x - 3)$$

B ভগ্নাংশের হর:

$$x^2 + x - 12 \\ = x^2 + 4x - 3x - 12 \\ = x(x + 4) - 3(x + 4) \\ = (x + 4)(x - 3)$$

C ভগ্নাংশের হর:

$$x^2 + 2x - 8 \\ = x^2 + 4x - 2x - 8 \\ = x(x + 4) - 2(x + 4) \\ = (x + 4)(x - 2)$$

ভগ্নাংশগুলোর হরের ল.সা.গু. = (x - 2)(x - 3)(x + 4)

$$\begin{gathered} \therefore A=\frac{1}{x^2-5x+6} \\ =\frac{1}{(x-2) (x-3)}=\frac{1\times (x+4)}{(x-2) (x-3) (x+4)} \end{gathered}$$

$B=\frac{2}{x^2+x-12}=\frac{2}{(x-3) (x+4)}=\frac{2\times (x-2)}{(x-2) (x-3) (x+4)}$

$$C = \frac{3}{x^2 + 2x - 8} = \frac{3}{(x - 2)(x + 4)} = \frac{3 \times (x - 3)}{(x - 2)(x - 3)(x + 4)}$$

নির্ণেয় সাধারণ হরবিশিষ্ট ভগ্নাংশগুলো হলো: $\frac{x+4}{(x-2) (x-3) (x+4)},\frac{2(x-2)}{(x-2) (x-3) (x+4)} \text{এবং }\frac{3(x-3)}{(x-2) (x-3) (x+4)}$

এখানে,

$$A = \frac{1}{x^2 - 5x + 6} = \frac{1}{(x-2)(x-3)}$$

$$B = \frac{2}{x^2 + x - 12} = \frac{2}{(x-3)(x+4)}$$

$$C = \frac{3}{x^2 + 2x - 8} = \frac{3}{(x-2)(x+4)}$$

$$\therefore A + B - C = \frac{1}{(x-2)(x-3)} + \frac{2}{(x-3)(x+4)} - \frac{3}{(x-2)(x+4)}$$

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

$$= \frac{x + 4 + 2x - 4 - 3x + 9}{(x-2)(x-3)(x+4)} = \frac{9}{(x-2)(x-3)(x+4)}$$

নির্ণেয় সরলফল: $$\frac{9}{(x-2)(x-3)(x+4)}$$