cm đẳng thức sau` (a^2 + 3ab) /( a^2 -9b^2) + (2a^2 - 5ab - 3b^2) / (6ab - a^2 -9b^2) = (a^2 - an + bn +ab) / (3bn -a^2 - an + 3ab)`

ĐKXĐ : $a\ne ±3b, a\ne -n$

$\dfrac{a^2+3ab}{a^2-9b^2}+\dfrac{2a^2-5ab-3b^2}{6ab-a^2-9b^2}$

$=\dfrac{a(a+3b)}{(a-3b)(a+3b)}-\dfrac{(a-3b)(2a+b)}{(a-3b)^2}$

$=\dfrac{a}{a-3b}-\dfrac{2a+b}{a-3b}$

$=\dfrac{a-2a-b}{a-3b}$

$=\dfrac{-a-b}{a-3b}$

$=\dfrac{a+b}{3b-a}(1)$

$\dfrac{a^2+an+bn+ab}{3bn-a^2-an+3ab}$

$=\dfrac{a(a+n)+b(a+n)}{n(3b-a)+a(3b-a)}$

$=\dfrac{(a+n)(a+b)}{(a+n)(3b-a)}$

$=\dfrac{a+b}{3b-a}(2)$

$(1),(2)=>\dfrac{a^2+3ab}{a^2-9b^2}+\dfrac{2a^2-5ab-3b^2}{6ab-a^2-9b^2}=\dfrac{a^2+an+bn+ab}{3bn-a^2-an+3ab}$