$\frac{1}{a-b},\frac{1}{a^2+ab+b^{}}$এই ভগ্নাংশদ্বয়ের যোগফল নির্ণয়  কর।

(সংক্ষিপ্ত প্রশ্ন)

$\frac{x^3{y}^3+x^2{y}^3}{x^2+xy}=\frac{x^2{y}^2(x+y)}{x(x+y)}=x{y}^2$

$\frac{x^2-y^2}{x^3-y^3} = \frac{(x+y)(x-y)}{(x-y)(x^2+xy+y^2)} = \frac{x+y}{x^2+xy+y^2}$

$$\begin{gathered} \frac{x^{3}y - x{y}^3}{x^3+y^3} = \frac{xy(x^2-y^2)}{(x+y)(x^2-xy+y^2)} \\ = \frac{xy(x+y)(x-y)}{(x+y)(x^2-xy-y^2)} = \frac{xy(x-y)}{x^2-xy+y^2} \end{gathered}$$

$$\begin{gathered} \frac{a^2+3a-10}{a^2-25} \\ =\frac{a^2+5a-2a-10}{{\left(a\right)}^2{\left(5\right)}^2} \\ =\frac{a(a+5) - 2 (a+5)}{(a+5)(a-5)} \\ =\frac{(a+5)(a-2)}{(a+5)(a-5)} = \frac{a-2}{a-5} \end{gathered}$$

$$\begin{gathered} \frac{8a^3 - 27b^3}{4a^{3}b + 6a^2{b}^2+9a{b}^3} \\ =\frac{{\left(2a\right)}^3- {\left(3b\right)}^3}{ab(4a^2 +6ab+9b^2)} \\ =\frac{(2a-3b) \{{\left(2a\right)}^2+2a3b+{\left(3b\right)}^2\}}{ab(4a^2+6ab+9b^2)} \\ =\frac{2a-3b}{ab} \end{gathered}$$

প্রদত্ত ভগ্নাংশ : $\frac{p+q}{p^3{q}^2}, \frac{p-q}{p^2{q}^3}$

এখানে হর p³q² ও p²q³   এর লসাগু = p³q³

১ম ভগ্নাংশ = $\frac{p+q}{p^2{q}^2}=\frac{p(p-q)}{p^3{q}^2\times q}=\frac{q(p+q)}{p^3{q}^3}$

২য় ভগ্নাংশ = $\frac{p-q}{p^2{q}^3}=\frac{p(p-q)}{p^2{q}^3\times q}=\frac{q(p+q)}{p^3{q}^3}$