Mi aiutate x cortesia a risolvere es. N. 268 grazie

(19/3 : (30+9-20)/24)-1/5) : x = x : ((1/2+4/25)-2/5)*10/3

(19/3*24/19-1/5) : x = x : ((25+8)/50-2/5)*10/3

(8-1/5) : x = x : 13/50*10/3

39/5 : x = x : 13/15

x^2 = 13*39/(15*5) = 169/25

x = 13/5

268)

$\small \left[\dfrac{19}{3} : \left(\dfrac{5}{4}+\dfrac{3}{8}-\dfrac{5}{6}\right)-\dfrac{1}{5}\right] : x = x : \left\{\left[\left(\dfrac{1}{2}+\dfrac{2}{\cancel{15}_5}·\dfrac{\cancel6^2}{5}\right)-\dfrac{2}{5}\right]·\dfrac{10}{3}\right\}$

$\small \left[\dfrac{19}{3} : \left(\dfrac{30+9-20}{24}\right)-\dfrac{1}{5}\right] : x = x : \left\{\left[\left(\dfrac{1}{2}+\dfrac{2}{5}·\dfrac{2}{5}\right)-\dfrac{2}{5}\right]·\dfrac{10}{3}\right\}$

$\small \left[\dfrac{19}{3} : \dfrac{19}{24}-\dfrac{1}{5}\right] : x = x : \left\{\left[\left(\dfrac{1}{2}+\dfrac{4}{25}\right)-\dfrac{2}{5}\right]·\dfrac{10}{3}\right\}$

$\small \left[\dfrac{\cancel{19}^1}{\cancel3_1}·\dfrac{\cancel{24}^8}{\cancel{19}_1}-\dfrac{1}{5}\right] : x = x : \left\{\left[\left(\dfrac{25+8}{50}\right)-\dfrac{2}{5}\right]·\dfrac{10}{3}\right\}$

$\small \left[\dfrac{1}{1}·\dfrac{8}{1}-\dfrac{1}{5}\right] : x = x : \left\{\left[\dfrac{33}{50}-\dfrac{2}{5}\right]·\dfrac{10}{3}\right\}$

$\small \left[8-\dfrac{1}{5}\right] : x = x : \left\{\left[\dfrac{33-20}{50}\right]·\dfrac{10}{3}\right\}$

$\small \left[\dfrac{40-1}{5}\right] : x = x : \left\{\dfrac{13}{\cancel{50}_5}·\dfrac{\cancel{10}^1}{3}\right\}$

$\small \dfrac{39}{5} : x = x : \left\{\dfrac{13}{5}·\dfrac{1}{3}\right\}$

$\small  \dfrac{39}{5} : x = x : \dfrac{13}{15}$

moltiplica tra loro in medi e gli estremi:

$\small  x·x = \dfrac{13}{\cancel{15}_5}·\dfrac{\cancel{39}^{13}}{5}$

$\small  x^2 = \dfrac{13}{5}·\dfrac{13}{5}$

$\small  x^2 = \dfrac{169}{25} $

radice di ambo le parti:

$\small  \sqrt{x^2} = \sqrt{\dfrac{169}{25}} $

$\small x = \dfrac{13}{5} $