Buongiorno a tutti e buon inizio settimana! Vi auguro che queste vacanze di Natale stiano procedendo bene! Ringrazio anticipatamente chi mi potrà aiutare a risolvere questo esercizio di algebra!

((- a^3·b + 8/5·a^3·b)·(- 10/9·a·b^3)/(+ 4/9·a^2·b^2) +

+(- 1/3·a·b)^2)^2/(- 25/27·a^2·b^2) =

=(3·a^3·b/5·(- 10/9·a·b^3)/(+ 4/9·a^2·b^2) +

+a^2·b^2/9)^2/(- 25/27·a^2·b^2) =

=((- 2·a^4·b^4/3)/(+ 4/9·a^2·b^2) +

+a^2·b^2/9)^2/(- 25/27·a^2·b^2) =

=(- 3·a^2·b^2/2 + a^2·b^2/9)^2/(- 25/27·a^2·b^2)=

=(- 25·a^2·b^2/18)^2/(- 25/27·a^2·b^2)=

=625·a^4·b^4/324/(- 25/27·a^2·b^2)=

=- 25·a^2·b^2/12

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$\small \left[\left(-a^3b+\dfrac{8}{5}a^3b\right)·\left(-\dfrac{10}{9}ab^3\right) : \left(+\dfrac{4}{9}a^2b^2\right)+\left(-\dfrac{1}{3}ab\right)^2\right]^2 : \left(-\dfrac{25}{27}a^2b^2\right)=$

$\small =\left[\left(\dfrac{-5+8}{5}a^3b\right)·\left(-\dfrac{10}{9}ab^3\right) : \dfrac{4}{9}a^2b^2 +\dfrac{1}{9}a^2b^2\right]^2 : \left(-\dfrac{25}{27}a^2b^2\right)=$

$\small =\left[\dfrac{3}{5}a^3b·\left(-\dfrac{10}{9}ab^3\right) : \dfrac{4}{9}a^2b^2 +\dfrac{1}{9}a^2b^2\right]^2 : \left(-\dfrac{25}{27}a^2b^2\right)=$

$\small =\left[-\dfrac{30}{45}a^{3+1}b^{1+3} : \dfrac{4}{9}a^2b^2 +\dfrac{1}{9}a^2b^2\right]^2 : \left(-\dfrac{25}{27}a^2b^2\right)=$

$\small =\left[-\dfrac{\cancel{30}^2}{\cancel{45}_3}a^4b^4 : \dfrac{4}{9}a^2b^2 +\dfrac{1}{9}a^2b^2\right]^2 : \left(-\dfrac{25}{27}a^2b^2\right)=$

$\small =\left[-\dfrac{2}{3}a^4b^4 : \dfrac{4}{9}a^2b^2 +\dfrac{1}{9}a^2b^2\right]^2 : \left(-\dfrac{25}{27}a^2b^2\right)=$

$\small =\left[-\dfrac{2}{3} ·\dfrac{9}{4}·\dfrac{a^4b^4}{a^2b^2} +\dfrac{1}{9}a^2b^2\right]^2 : \left(-\dfrac{25}{27}a^2b^2\right)=$

$\small =\left[-\dfrac{\cancel2^1}{\cancel3_1} ·\dfrac{\cancel9^3}{\cancel4_2}·\dfrac{a^{\cancel4^2}b^{\cancel4^2}}{\cancel{a^2}\cancel{b^2}} +\dfrac{1}{9}a^2b^2\right]^2 : \left(-\dfrac{25}{27}a^2b^2\right)=$

$\small =\left[-1 ·\dfrac{3}{2}a^2b^2 +\dfrac{1}{9}a^2b^2\right]^2 : \left(-\dfrac{25}{27}a^2b^2\right)=$

$\small =\left[-\dfrac{3}{2}a^2b^2 +\dfrac{1}{9}a^2b^2\right]^2 : \left(-\dfrac{25}{27}a^2b^2\right)=$

$\small =\left[\dfrac{-27+2}{18}a^2b^2\right]^2 : \left(-\dfrac{25}{27}a^2b^2\right)=$

$\small =\left[-\dfrac{25}{18}a^2b^2\right]^2 : \left(-\dfrac{25}{27}a^2b^2\right)=$

$\small=\dfrac{25^2}{18^2}a^4b^4 : -\dfrac{25}{27}a^2b^2=$

$\small=\dfrac{\cancel{625}^{25}}{\cancel{324}_{12}}· -\dfrac{\cancel{27}^1}{\cancel{25}_1}\dfrac{a^{\cancel4^2}b^{\cancel4^2} }{\cancel{a^2}\cancel{b^2}}=$

$\small=\dfrac{25}{12}· -\dfrac{1}{1}a^2b^2= -\dfrac{25}{12}a^2b^2$