[Risolto] Seconda media

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$\left(x+\dfrac{3}{4}\right) : x = \left(\dfrac{5}{2}+5\right) : \left(1+\dfrac{7}{3}\right)$

$\left(x+\dfrac{3}{4}\right) : x = \left(\dfrac{5+10}{2}\right) : \left(\dfrac{3+7}{3}\right)$

$\left(x+\dfrac{3}{4}\right) : x = \dfrac{15}{2} : \dfrac{10}{3}$

applica lo scomporre:

$\left(x+\dfrac{3}{4}-x\right) : x = \left(\dfrac{15}{2}-\dfrac{10}{3}\right) : \dfrac{10}{3}$

$\dfrac{3}{4} : x = \left(\dfrac{45-20}{6}\right) : \dfrac{10}{3}$

$\dfrac{3}{4} : x = \dfrac{25}{6} : \dfrac{10}{3}$

$x= \dfrac{\cancel3^1}{\cancel4_2}·\dfrac{\cancel{10}^5}{\cancel3_1} : \dfrac{25}{6}$

$x= \dfrac{\cancel5^1}{\cancel2_1}·\dfrac{\cancel6^3}{\cancel{25}_5}$

$x= \dfrac{3}{5}$

prodotto dei medi è uguale al prodotto degli estremi:

$x=(x+3/4)*(1+7/3)/(5/2+5)$
$x=(x+3/4)*(10/3)/(15/2)$
$x=(10/3x + 5/2)/15/2$
$x=(20x+15)/45$
$x=(4x+3)/9$
$9x=4x+3$
$5x=3$
$x=3/5$

(4x+3)/4 : x = 15/2 : 10/3

si fa il prodotto in croce :

(4x+3)/4 * 10/3  = 15x/2

(40x+30)/12 = 15x/2 

(40x+30)/12= 15x*6/12 

30 = 50 x

x = 3/5