Aspressione con frazioni

Prima parentesi quadrata:

[(1 - 1/2)^3 * (- 1/3 - 5) + 1/6]^3....

[(2/2 - 1/2)^3 * (- 1/3 - 15/3) + 1/6]^3....

[(1/2)^3 * (- 16/3) + 1/6]^3....

[(1/8) * (- 16/3) + 1/6]^3....

[- 2/3 + 1/6]^3 ...

[- 4/6 + 1/6]^3...

[-3/6]^3...

[- 1/2]^3 ...

- 1/8....; continua...

aggiungiamo la seconda parentesi quadrata e tutto il resto:

- 1/8 + [(2/9 - 1/15 + 7/5): 56/33] *  (- 3^2) - (- 1/2)^2 + (4 + 2/3 : 4/9);

mcm(9; 15; 5)  = 9 * 5 = 45;        mcm(1; 3; 9) = 9;

- 1/8 + [(10/45 - 3/45 + 63/45) : 56/33] * (- 9) - (+ 1/4) + (4 + 2/3 *  9/4)=

= - 1/8 + [70/45 : 56/33] * (- 9) - 1/4 + (4 + 3/2) = 

= - 1/8 + [14/9 * 33/56] * (- 9) - 1/4 + ( 8/2 + 3/2)=;   (56 si semplifica per 14; 56:14 = 4;  33 si divide per 3)

= - 1/8 + [1 * 11/(3 * 4)] * (- 9) - 1/4 + 11/2 =

= - 1/8 + [11/12] * (- 9) - 1/4 + 11/2 =  ;      (11 * (-  9) /12 = - 33/4  ; 12 si semplifica con il 9 per 3 ).

= - 1/8 - 33/4 - 1/4 + 11/2 =

mcm  = 8;

= - 1/8 - 66/8 - 2/8 + 44/8 =

= - 25/8.

Ciao  @fabroxy

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218)

$\small \left[\left(1-\dfrac{1}{2}\right)^3·\left(-\dfrac{1}{3}-5\right)+\dfrac{1}{6}\right]^3+\left[\left(\dfrac{2}{9}-\dfrac{1}{15}+\dfrac{7}{5}\right) : \dfrac{56}{33}\right]·\left(-3^2\right)-\left(-\dfrac{1}{2}\right)^2+\left(4+\dfrac{2}{3} : \dfrac{4}{9}\right) =$

$\small =\left[\left(\dfrac{2-1}{2}\right)^3·\left(\dfrac{-1-15}{3}\right)+\dfrac{1}{6}\right]^3+\left[\left(\dfrac{10-3+63}{45}\right)·\dfrac{33}{56}\right]·-9-\dfrac{1}{4}+\left(4+\dfrac{\cancel2^1}{\cancel3_1}·\dfrac{\cancel9^3}{\cancel4_2}\right) =$

$\small =\left[\left(\dfrac{1}{2}\right)^3·-\dfrac{16}{3}+\dfrac{1}{6}\right]^3+\left[\dfrac{\cancel{70}^{14}}{\cancel{45}_9}·\dfrac{33}{56}\right]·-9-\dfrac{1}{4}+\left(4+\dfrac{1}{1}·\dfrac{3}{2}\right) =$

$\small =\left[\dfrac{1}{\cancel8_1}·-\dfrac{\cancel{16}^2}{3}+\dfrac{1}{6}\right]^3+\left[\dfrac{\cancel{14}^1}{\cancel9_3}·\dfrac{\cancel{33}^{11}}{\cancel{56}_4}\right]·-9-\dfrac{1}{4}+\left(4+\dfrac{3}{2}\right) =$

$\small =\left[\dfrac{1}{1}·-\dfrac{2}{3}+\dfrac{1}{6}\right]^3+\left[\dfrac{1}{3}·\dfrac{11}{4}\right]·-9-\dfrac{1}{4}+\left(\dfrac{8+3}{2}\right) =$

$\small =\left[-\dfrac{2}{3}+\dfrac{1}{6}\right]^3+\dfrac{11}{\cancel{12}_4}·-\cancel9^3-\dfrac{1}{4}+\dfrac{11}{2} =$

$\small =\left[\dfrac{-4+1}{6}\right]^3+\dfrac{11}{4}·-3-\dfrac{1}{4}+\dfrac{11}{2} =$

$\small =\left[-\dfrac{\cancel3^1}{\cancel6_2}\right]^3-\dfrac{33}{4}-\dfrac{1}{4}+\dfrac{11}{2} =$

$\small =\left[-\dfrac{1}{2}\right]^3-\dfrac{\cancel{34}^{17}}{\cancel4_2}+\dfrac{11}{2} =$

$\small =-\dfrac{1}{8}-\dfrac{17}{2}+\dfrac{11}{2} =$

$\small =\dfrac{-1-68+44}{8}=$

$\small = -\dfrac{25}{8}$