[Risolto] emergenza

3)

((1/4 : 5/4 +1)alla seconda : 9/5) : x = x : ((5/2 - 1/3 - 7/4) x 9/5) : 

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$\big[\big(\frac{1}{4} : \frac{5}{4} +1\big)^2 : \frac{9}{5}\big] : x = x : \big[\big(\frac{5}{2}-\frac{1}{3}-\frac{7}{4}\big)·\frac{9}{5}\big]$

$\big[\big(\frac{1}{4}·\frac{4}{5} +1\big)^2·\frac{5}{9}\big] : x = x : \big[\big(\frac{30-4-21}{12}\big)·\frac{9}{5}\big]$

$\big[\big(\frac{1}{1}·\frac{1}{5} +1\big)^2·\frac{5}{9}\big] : x = x : \big[\frac{5}{12}·\frac{9}{5}\big]$

$\big[\big(\frac{1}{5} +1\big)^2·\frac{5}{9}\big] : x = x : \big[\frac{1}{4}·\frac{3}{1}\big]$

$\big[\big(\frac{1+5}{5}\big)^2·\frac{5}{9}\big] : x = x : \frac{3}{4}$

$\big[\big(\frac{6}{5}\big)^2·\frac{5}{9}\big] : x = x : \frac{3}{4}$

$\big[\frac{36}{25}·\frac{5}{9}\big] : x = x : \frac{3}{4}$

$\big[\frac{4}{5}·\frac{1}{1}\big] : x = x : \frac{3}{4}$

$\frac{4}{5} : x = x : \frac{3}{4}$

$x^2 = \frac{3}{4}·\frac{4}{5}$

$x^2 = \frac{3}{1}·\frac{1}{5}$

$x^2 = \frac{3}{5}$

$\sqrt{x^2} = \sqrt{\frac{3}{5}}$

$x= ±\frac{\sqrt{15}}{5} $ → $~(≅ ±0,774597)$.

Scusa, ma non si capisce niente. Poi un esercizio per volta come da:

https://www.sosmatematica.it/regolamento/

palesando le tue difficoltà risolutive relative all'esercizio svolto.