Ho delle difficoltà

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$\small \left[\left(2^8x^4y^7\right)÷\left(2^6x^2y^4\right)+\left(2^6x^4y^3\right)^2÷\left(2^{14}x^6y^3\right)\right]÷\left[\left(\dfrac{2}{5}\right)^{-2}x^2y\right]+2\left(-\dfrac{2}{5}y\right)^2 =$

$\small =\dfrac{\dfrac{2^8x^4y^7}{2^6x^2y^4}+\dfrac{\left(2^6x^4y^3\right)^2}{2^{14}x^6y^3}}{{\left(\dfrac{2}{5}\right)^{-2}x^2y}}+2\left(-\dfrac{2}{5}y\right)^2 =$

$\small =\dfrac{2^{8-6}x^{4-2}y^{7-4}+\dfrac{2^{6·2}x^{4·2}y^{3·2}}{2^{14}x^6y^3}}{\left(\dfrac{5}{2}\right)^2x^2y}+2·\dfrac{4}{25}y^2 =$

$\small =\dfrac{2^2x^2y^3+\dfrac{2^{12}x^8y^6}{2^{14}x^6y^3}}{\dfrac{25}{4}x^2y}+\dfrac{8}{25}y^2 =$

$\small =\dfrac{2^2x^2y^3+2^{12-14}x^{8-6}y^{6-3}}{\dfrac{25}{4}x^2y}+\dfrac{8}{25}y^2 =$

$\small =\dfrac{2^2x^2y^3+2^{-2}x^2y^3}{\dfrac{25}{4}x^2y}+\dfrac{8}{25}y^2 =$

$\small =\dfrac{4x^2y^3+\left(\dfrac{1}{2}\right)^2x^2y^3}{\dfrac{25}{4}x^2y}+\dfrac{8}{25}y^2 =$

$\small =\dfrac{4x^2y^3+\dfrac{1}{4}x^2y^3}{\dfrac{25}{4}x^2y}+\dfrac{8}{25}y^2 =$

$\small =\dfrac{\dfrac{16+1}{4}x^2y^3}{\dfrac{25}{4}x^2y}+\dfrac{8}{25}y^2 =$

$\small =\dfrac{\dfrac{17}{4}x^2y^3}{\dfrac{25}{4}x^2y}+\dfrac{8}{25}y^2 =$

$\small = \dfrac{17}{\cancel4_1}x^2y^3·\dfrac{\cancel4^1}{25}x^{-2}y^{-1}+\dfrac{8}{25}y^2 =$

$\small = \dfrac{17}{25}x^{2+(-2)}y^{3+(-1)}+\dfrac{8}{25}y^2 =$

$\small = \dfrac{17}{25}x^{2-2}y^{3-1}+\dfrac{8}{25}y^2 =$

$\small = \dfrac{17}{25}x^0y^2+\dfrac{8}{25}y^2 =$

$\small = \dfrac{17}{25}·1y^2+\dfrac{8}{25}y^2 =$

$\small = \dfrac{17}{25}y^2+\dfrac{8}{25}y^2 =$

$\small = \dfrac{\cancel{25}^1}{\cancel{25}_1}y^2 =$

$\small = y^2 $

2^8 x^4 y^2 : (2^6 x^2 y^4) = 2^2 x^4 /y^2 = 4 x^4 y^-2