Mate

Riguarda il testo. Non si legge molto bene ...

(- 1/3)^3 = - 1/27;

Tutte le basi elevate a esponente 0 danno 1.

1/2 - (- 1/27) +{[(- 12/3 - 4/3) * (+ 3/16) + 1/5]^2 - (- 1/9 - 9/9)}^0.... 

Il risultato della parentesi graffa è 1;

{[(- 12/3 - 4/3) * (+ 3/16) + 1/5]^2 - (- 1/9 - 9/9)}^0 = 1

ciao @madalina87

prima:

$$\frac{1}{2} - \left(-\frac{1}{3}\right)^3 + \left\{ \left[ \left(-4 - \frac{4}{3}\right) : \left(+\frac{16}{3}\right) + \frac{1}{5} \right]^2 - \left(-\frac{1}{9} - 1\right) \right\}^0 \cdot \left(\frac{1}{9} - \frac{7}{15} + \frac{1}{45}\right)^3 - \frac{2}{3}$$

$$\frac{1}{2} - \left(-\frac{1}{27}\right) + \left\{ \left[ \left(-\frac{16}{3}\right) : \left(\frac{16}{3}\right) + \frac{1}{5} \right]^2 - \left(-\frac{10}{9}\right) \right\}^0 \cdot \left(-\frac{15}{45}\right)^3 - \frac{2}{3}$$
$$\frac{1}{2} + \frac{1}{27} + \left\{ \left[ -1 + \frac{1}{5} \right]^2 + \frac{10}{9} \right\}^0 \cdot \left(-\frac{1}{3}\right)^3 - \frac{2}{3}$$
$$\frac{1}{2} + \frac{1}{27} + \left\{ \left[ -\frac{4}{5} \right]^2 + \frac{10}{9} \right\}^0 \cdot \left(-\frac{1}{27}\right) - \frac{2}{3}$$
$$\frac{1}{2} + \frac{1}{27} + \left\{ \frac{16}{25} + \frac{10}{9} \right\}^0 \cdot \left(-\frac{1}{27}\right) - \frac{2}{3}$$
$$\frac{1}{2} + \frac{1}{27} + 1 \cdot \left(-\frac{1}{27}\right) - \frac{2}{3}$$
$$\frac{1}{2} + \frac{1}{27} - \frac{1}{27} - \frac{2}{3}$$
$$\frac{1}{2} - \frac{2}{3} = \frac{3 - 4}{6} = -\frac{1}{6}$$

seconda:
$$\left[ \left( 1 + \frac{1}{2} \right) \cdot \left( \frac{3}{2} \right)^{-1} - \left( 2 - \frac{1}{3} \right) \cdot \left( -\frac{5}{3} \right)^{-2} \cdot 3^{-1} \right] \cdot \frac{4}{5}$$

$$\left[ \left( \frac{3}{2} \right) \cdot \frac{2}{3} - \left( \frac{5}{3} \right) \cdot \left( -\frac{3}{5} \right)^{2} \cdot \frac{1}{3} \right] \cdot \frac{4}{5}$$
$$\left[ 1 - \frac{5}{3} \cdot \frac{9}{25} \cdot \frac{1}{3} \right] \cdot \frac{4}{5}$$
$$\left[ 1 - \frac{45}{225} \right] \cdot \frac{4}{5}$$
$$\left[ 1 - \frac{1}{5} \right] \cdot \frac{4}{5}$$
$$\frac{4}{5} \cdot \frac{4}{5} = \frac{16}{25}$$

1/2 - (- 1/3)^3 + (((-4 - 4/3)/(+ 16/3) + 1/5)^2 - (- 1/9 - 1))^0·(1/9 - 7/15 + 1/45)^3 - 2/3=

=1/2 + 1/27 + (((-4 - 4/3)/(+ 16/3) + 1/5)^2 - (- 1/9 - 1))^0·(1/9 - 7/15 + 1/45)^3 - 2/3=

=1/2 + 1/27 + (((- 16/3)/(+ 16/3) + 1/5)^2 + 10/9)^0·(- 1/3)^3 - 2/3=

=1/2 + 1/27 + ((-1 + 1/5)^2 + 10/9)^0·(- 1/3)^3 - 2/3=

=1/2 + 1/27 + ((- 4/5)^2 + 10/9)^0·(- 1/3)^3 - 2/3=

=1/2 + 1/27 + (16/25 + 10/9)^0·(- 1/3)^3 - 2/3=

=1/2 + 1/27 + (394/225)^0·(- 1/3)^3 - 2/3=

=1/2 + 1/27 + 1·(- 1/3)^3 - 2/3=

=1/2 + 1/27 - 1/27 - 2/3 = - 1/6

((1 + 1/2)·(3/2)^(-1) - (2 - 1/3)·(- 5/3)^(-2)·3^(-1))·4/5=

=(3/2·(2/3) - 5/3·(- 5/3)^(-2)·3^(-1))·4/5=

=(3/2·(2/3) - 5/3·(9/25)·(1/3))·4/5=

=(1 - 5/3·(9/25)·(1/3))·4/5=

=(1 - 1/5)·4/5=

=4/5·4/5 = 16/25

Non sei a scuola, vero ?

Comincio a guardarla e poi pubblico uno svolgimento dopo le 14.

$\small\text{1° espressione:}$

$\small \dfrac{1}{2}-\left(-\dfrac{1}{3}\right)^3+\left\{\left[\left(-4-\dfrac{4}{3}\right)÷\left(\dfrac{16}{3}\right)+\dfrac{1}{5}\right]^2-\left(-\dfrac{1}{9}-1\right)\right\}^0·\left(\dfrac{1}{9}-\dfrac{7}{15}+\dfrac{1}{45}\right)^3-\dfrac{2}{3} =$

$\small\text{ciò che è compreso nella graffa essendo elevato a zero diventa 1 per cui:}$

$\small = \dfrac{1}{2}-\left(-\dfrac{1}{27}\right)+1·\left(\dfrac{5-21+1}{45}\right)^3-\dfrac{2}{3} =$

$\small = \dfrac{1}{2}-\left(-\dfrac{1}{27}\right)+\left(-\dfrac{\cancel{15}^1}{\cancel{45}_3}\right)^3-\dfrac{2}{3} =$

$\small = \dfrac{1}{2}+\dfrac{1}{27}+\left(-\dfrac{1}{3}\right)^3-\dfrac{2}{3} =$

$\small = \dfrac{1}{2}\cancel{+\dfrac{1}{27}}\cancel{-\dfrac{1}{27}}-\dfrac{2}{3} =$

$\small = \dfrac{1}{2}-\dfrac{2}{3} =$

$\small = \dfrac{3-4}{6} =$

$\small = -\dfrac{1}{6} $

$\small\text{2° espressione:}$

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

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

$\small =\left[\dfrac{\cancel3^1}{\cancel2_1}·\dfrac{\cancel2^1}{\cancel3_1}-\dfrac{\cancel5^1}{\cancel3_1}·\dfrac{\cancel9^3}{\cancel{25}_5}·\dfrac{1}{3}\right]·\dfrac{4}{5} =$

$\small =\left[\dfrac{1}{1}·\dfrac{1}{1}-\dfrac{1}{1}·\dfrac{\cancel3^1}{5}·\dfrac{1}{\cancel3_1}\right]·\dfrac{4}{5} =$

$\small =\left[1-\dfrac{1}{1}·\dfrac{1}{5}·\dfrac{1}{1}\right]·\dfrac{4}{5} =$

$\small =\left[1-\dfrac{1}{5}\right]·\dfrac{4}{5} =$

$\small =\left[\dfrac{5-1}{5}\right]·\dfrac{4}{5} =$

$\small =\dfrac{4}{5}·\dfrac{4}{5} =$

$\small =\dfrac{16}{25} $