\(x=\dfrac{3a+3}{6a+15b}\) ; \(y=\dfrac{4a^2-25b^2}{a^3+1}\)
\(\Rightarrow A=\dfrac{x^2-x}{2}.\dfrac{2y}{x-1}=\dfrac{2xy\left(x-1\right)}{2\left(x-1\right)}=xy\)
\(\Rightarrow A=\dfrac{3\left(a+1\right)}{3\left(2a+5b\right)}.\dfrac{4a^2-25b^2}{a^3+1}=\dfrac{3\left(a+1\right)}{3\left(2a+5b\right)}.\dfrac{\left(2a+5b\right)\left(2a-5b\right)}{\left(a+1\right)\left(a^2-a+1\right)}\)
\(=\dfrac{2a-5b}{a^2-a+1}\)
Ta có:
\(\left(6a+15b\right)=3a+3\)
\(\Leftrightarrow x=\dfrac{3a+3}{6a+15b}=\dfrac{a+1}{2a+5b}\)
\(\left(a^3+1\right)y=4a^2-25b^2\Leftrightarrow y=\dfrac{4a^2-25b^2}{a^3+1}\)
\(A=\dfrac{x^2-x}{2}\cdot\dfrac{2y}{x-1}=\dfrac{x\left(x-1\right)}{2}\cdot\dfrac{2y}{x-1}=\dfrac{2xy}{2}=xy\)
\(=\dfrac{a+1}{2a+5b}\cdot\dfrac{4a^2-25^2}{a^3+1}=\dfrac{a+1}{2a+5b}\cdot\dfrac{\left(2a+5b\right)\left(2a-5b\right)}{\left(a+1\right)\left(a^2-a+1\right)}=\dfrac{2a-5n}{a^2-a+1}\)
