Espressioni calcolo letterale (3 media)..un aiuto per sapere almeno il risultato grazie!

((- 12·a + 5·a - 3·a)·(- 2/5·a·x) + 3/5·x·(3·a^2 - 7·a^2 - 6·a^2))/(- 8·a·x)=

=((- 10·a)·(- 2/5·a·x) + 3/5·x·(- 10·a^2))/(- 8·a·x)=

=(4·a^2·x + (- 6·a^2·x))/(- 8·a·x)=

=(- 2·a^2·x)/(- 8·a·x)=

=a/4

a ≠ 0 ∧ x ≠ 0

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(- ((- 3·a^2·b)^3 - 60·a^7·b^3/(- 20·a) + 6·a^5·(- 12·a·b^3 + 9·a·b^3))/(- 21·a^6))^3=

=(- (- 27·a^6·b^3 - (- 3·a^6·b^3) + 6·a^5·(- 3·a·b^3))/(- 21·a^6))^3=

=(- (- 27·a^6·b^3 + 3·a^6·b^3 - 18·a^6·b^3)/(- 21·a^6))^3=

=(- (- 42·a^6·b^3)/(- 21·a^6))^3=

=(- 2·b^3)^3 = - 8·b^9

a ≠ 0

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10/a)

$\small \dfrac{\left[\left(-12a+5a-3a\right)·\left(-\dfrac{2}{5}ax\right)+\dfrac{3}{5}x·\left(3a^2-7a^2-6a^2\right)\right]}{-8ax} =$

$\small =\dfrac{\left[-\cancel{10}^2a·\left(-\dfrac{2}{\cancel5_1}ax\right)+\dfrac{3}{\cancel5_1}x·\left(-\cancel{10}^2a^2\right)\right]}{-8ax}=$

$\small =\dfrac{-2a·(-2ax) + 3x·\left(-2a^2\right)}{-8ax}=$

$\small =\dfrac{4a^2x- 6a^2x}{-8ax}=$

$\small =\dfrac{- 2a^2x}{-8ax}=$

$\small =\dfrac{- \cancel2^1a^\cancel2\cancel{x}}{-\cancel8_4\cancel{a}\cancel{x}}=$

$\small =\dfrac{1}{4}a= \dfrac{a}{4}$

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10/b)

$\small \left\{-\left[\left(-3a^2b\right)^3 -60a^7b^3 : \left(-20a\right) +6a^5·\left(-12ab^3+9ab^3\right)\right] : \left(-21a^6\right)\right\}^3=$

$\small = \left \{-\left[-3^{1·3}a^{2·3}b^{1·3} + 3a^{7-1}b^3 +6a^5·\left(-3ab^3\right)\right] : \left(-21a^6\right) \right\}^3 =$

$\small = \left \{-\left[-3^3a^6b^3 + 3a^6b^3 -18a^{5+1}b^3\right] : \left(-21a^6\right) \right\}^3 =$

$\small = \left\{-\left[-27a^6b^3 + 3a^6b^3 -18a^6b^3\right] : \left(-21a^6\right) \right\}^3 =$

$\small = \left\{-\left[-42a^6b^3\right] : \left(-21a^6\right) \right\}^3 =$

$\small = \left\{42a^6b^3 : \left(-21a^6\right) \right\}^3 =$

$\small = \left\{-2a^{6-6}b^3 \right\}^3 =$

$\small = \left\{-2a^0b^3 \right\}^3 =$

$\small = \left\{-2·1·b^3 \right\}^3 =$

$\small = \left\{-2b^3 \right\}^3 =$

$\small = -8b^9 $