[Risolto] aiutatemi in questa!! graziee!

269)

$\big[\big(\frac{5}{11}×\frac{33}{4}-\frac{3}{17}×\frac{34}{5}\big) : \frac{5}{3}\big] : x = x : \big\{\big[\frac{13}{20}+\big(\frac{17}{4}-\frac{3}{5}-\frac{5}{6}\big)-\frac{7}{3}\big] : \frac{1}{15}\big\}$

$\big[\big(\frac{5}{1}×\frac{3}{4}-\frac{3}{1}×\frac{2}{5}\big)×\frac{3}{5}\big] : x = x : \big\{\big[\frac{13}{20}+\big(\frac{255-36-50}{60}\big)-\frac{7}{3}\big]×15\big\}$

$\big[\big(\frac{15}{4}-\frac{6}{5}\big)×\frac{3}{5}\big] : x = x : \big\{\big[\frac{13}{20}+\frac{169}{60}-\frac{7}{3}\big]×15\big\}$

$\big[\big(\frac{75-24}{20}\big)×\frac{3}{5}\big] : x = x : \big\{\big[\frac{39+169-140}{60}\big]×15\big\}$

$\big[\frac{51}{20}×\frac{3}{5}\big] : x = x : \big\{\frac{68}{60}×15\big\}$

$\frac{153}{100} : x = x : \big\{\frac{17}{15}×15\big\}$

$\frac{153}{100} : x = x : 17$ moltiplica tra loro i medi egli estremi:

$x^2 = \frac{153}{100}×17$

$x^2 = \frac{2601}{100}$ radice quadrata di ambo le parti:

$\sqrt{x^2} = \sqrt{\frac{2601}{100}}$

$x= \frac{51}{10}$.

Prima le moltiplicazioni:   5/11 * 33/4 = 5/1 * 3/4 = 15/4; 

3/17 * 34/5 = 3/1 * 2/5 = 6/5.

mcd(4; 5; 6) = 60.

mcd (4; 5) = 20.

[( 15/4 - 6/5) * 3/5] : x = x : {[13/20 + (17 * 15 / 60 - 3 * 12/60 - 5 * 10/60) - 7/3] * 15/1};

[(15 * 5/20 - 6 * 4 / 20) * 3/5] : x = x : {[13/20 + (255/60 - 36/60 - 50/60) - 7/3 ] * 15/1};

[(75/20 - 24/20) * 3/5] : x = x : { [13/20 + 169/60 - 7/3] * 15/1};

[51/20 *  3/5] : x = x : {[13 * 3/60 + 169/60 - 7 * 20/60] * 15/1};

153/100 : x = x : {[39/60 + 169/60 - 140/60] * 15/1};

153/100 : x = x : {68/60 * 15/1};

68/60 * 15/1 = 68/4 * 1/1 = 17/1

153/100 : x = x : 17/1;

x^2 = 153/100 * 17/1;

x^2 = 2601 /100;

x = radicequadrata(2601 / 100) = 51/10.

Ciao  @gaiasonoio