Monomi234ho difficoltà

- (13 - 1)/9·x·(9·x) + (0.8·x)·(25/8·x) - 1/2·x^2 - (- 1/2·x^2)·(6/9·x) - (23 - 2)/9·x^3=

=- 4/3·x·(9·x) + 4·x/5·(25/8·x) - 1/2·x^2 - - x^3/3 - 7·x^3/3=

=(- 12·x^2 + 5·x^2/2 - 1/2·x^2) + x^3/3 - 7·x^3/3=

=- 2·x^3 - 10·x^2

coefficienti periodici:

1,3 (3 periodico) = (13 - 1) / 9 = 12/9;

12/9 = 4/3;

- (4/3) x ;

0,6 (6 periodico) = 6/9 = 2/3

(2/3) x;

2,3(3periodico) = (23 - 2)/9 = 21/9;

21/9 = 7/3;

(7/3) x^3.

Ciao @ozuna

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$\small \left(-1,\overline3x\right)(9x)+(0,8x)\left(\dfrac{25}{8}x\right)-\dfrac{1}{2}x^2-\left(-\dfrac{1}{2}x^2\right)\left(0,\overline6x\right)-2,\overline3x^3 =$

$\small = \left(-\dfrac{13-1}{9}x\right)·9x+\dfrac{\cancel8^4}{\cancel{10}_5}x·\dfrac{25}{8}x-\dfrac{1}{2}x^2-\left(-\dfrac{1}{2}x^2\right)\left(\dfrac{6-0}{9}x\right)-\dfrac{23-2}{9}x^3 =$

$\small = \left(-\dfrac{\cancel{12}^4}{\cancel9_3}x\right)·9x+\dfrac{\cancel4^1}{\cancel5_1}x·\dfrac{\cancel{25}^5}{\cancel8_2}x-\dfrac{1}{2}x^2-\left(-\dfrac{1}{2}x^2\right)\left(\dfrac{\cancel6^2}{\cancel9_3}x\right)-\dfrac{\cancel{21}^7}{\cancel9_3}x^3 =$

$\small = -\dfrac{4}{\cancel3_1}x·\cancel9^3x+x·\dfrac{5}{2}x-\dfrac{1}{2}x^2-\left(-\dfrac{1}{\cancel2_1}x^2\right)·\dfrac{\cancel2^1}{3}x-\dfrac{7}{3}x^3 =$

$\small = -4x·3x+\dfrac{5}{2}x^2-\dfrac{1}{2}x^2-\left(-x^2\right)·\dfrac{1}{3}x-\dfrac{7}{3}x^3 =$

$\small = -12x^2+\dfrac{\cancel4^2}{\cancel2_1}x^2-\left(-\dfrac{1}{3}x^3\right)-\dfrac{7}{3}x^3 =$

$\small = -12x^2+2x^2+\dfrac{1}{3}x^3-\dfrac{7}{3}x^3 =$

$\small = -10x^2-\dfrac{\cancel6^2}{\cancel3_1}x^3 =$

$\small = -10x^2-2x^3 =$

riordina:

$\small = -2x^3-10x^2$