radici quadrate ed espressioni

Prima risolvi l'espressione sotto radice; poi estrai la radice quadrata del risultato ottenuto!

{[1/2 x (5/4 - 2/4)]^2 x [(3/6 + 2/6 - 1/6)^2 : (1/6 + 3/6)^2] x (3/3 + 1/3) +

+ (32/16 - 3/16)} : 1/2 =

=  {[1/2 x 3/4]^2 x [(4/6)^2 : (4/6)^2] x (4/3) + (29/16)} : 1/2 =

= {[3/8]^2 x [2/3^2 : (2/3)^2] x (4/3) + (29/16)} : 1/2 =

= {9/64  x  [2/3]^0 x (4/3) + (29/16)} : 1/2 =

= {9/64 x 1 x 4/3 + 29/16} : 1/2 =

= {3/16 + 29/16} : 1/2 =

= 32/16 : 1/2 =

= 2/1 x 2/1 = 4. Ora estraiamo la radice di 4;

radicequadrata(4) = 2.

Ciao  @chiara0116

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$\small \sqrt{\left\{\left[\dfrac{1}{2}·\left(\dfrac{5}{4}-\dfrac{1}{2}\right)\right]^2·\left[\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{6}\right)^2 : \left(\dfrac{1}{6}+\dfrac{1}{2}\right)^2\right]·\left(1+\dfrac{1}{3}\right)+\left(2-\dfrac{3}{16}\right) \right\} : \dfrac{1}{2}} =$

$\small =\sqrt{\left\{\left[\dfrac{1}{2}·\left(\dfrac{5-2}{4}\right)\right]^2·\left[\left(\dfrac{3+2-1}{6}\right)^2 : \left(\dfrac{1+3}{6}\right)^2\right]·\left(\dfrac{3+1}{3}\right)+\left(\dfrac{32-3}{16}\right) \right\}·2} =$

$\small = \sqrt{\left\{\left[\dfrac{1}{2}·\dfrac{3}{4}\right]^2·\left[\left(\dfrac{4}{6}\right)^2 : \left(\dfrac{4}{6}\right)^2\right]·\dfrac{4}{3}+\dfrac{29}{16} \right\}·2} =$

$\small = \sqrt{\left\{\left[\dfrac{3}{8}\right]^2·1·\dfrac{4}{3}+\dfrac{29}{16} \right\}·2} =$

$\small = \sqrt{\left\{\dfrac{\cancel9^3}{\cancel{64}_{16}}·\dfrac{\cancel4^1}{\cancel3_1}+\dfrac{29}{16} \right\}·2} =$

$\small = \sqrt{\left\{\dfrac{3}{16}+\dfrac{29}{16} \right\}·2} =$

$\small = \sqrt{\dfrac{32}{\cancel{16}_8} ·\cancel2^1} =$

$\small = \sqrt{\dfrac{32}{8} } =$

$\small = \sqrt{4} = 2$