Numeri decimali e frazioni

Foto dritta: un esercizio per volta!

((2/3)^5·(2/3)^3)^2/((2/3)^2)^7 - 4/10/(4/9)·(2/3/(3/5))·(1/9)=

=(2^8/3^8)^2/(2^2/3^2)^7 - 9/10·(10/9)·(1/9)=

=2^16/3^16/(2^14/3^14) - 1/9=

=4/9 - 1/9= 3/9 = 1/3

La seconda:

0,49 (con 49 periodico) = (49 - 0) /99 = 49/99 ;

(- 7/3)^(-2) = (- 3/7) ^(+2)

49/99 : (- 7/3)^2; si trasforma in moltiplicazione con la frazione inversa;

49/99 * (- 3/7)^2 = 49/99 * (+ 9/49) = (49 * 9) / 99 * 49) = 9/99 =  1/11;

1/11 - (8/3 : 11/9) * [(1/10)^-1 : (- 6/10)^-1] =

= 1/11 - (8/3 * 9/11) * [(10/1) :  (- 10/6)] =

1/11 - (8 * 9)/(3 * 11) * [(10/1) :  (- 5/3)] =

= 1/11 - 24/11 * [10/1 * (- 3/5] =

= 1/11 - 24/11 *  [2 *(-3)] =

= 1/11 - 24/11 * (- 6) = 

= 1/11 + (24 * 6) / 11 =

= 1/11 + 144/11 =  145/11.

Ciao  @pia_maria

((2/3)^((5+3)*2) / (2/3)^(2*7) - 9/10 * 10/9 * 1/9

2/3^2 - 1/9

4/9 - 1/9 

3/9

1/3

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$\small\left[\left(0,\overline6\right)^5·\left(0,\overline6\right)^3\right]^2÷\left[\left(0,\overline6\right)^2\right]^7-\dfrac{0,4}{0,\overline4}·\dfrac{0,\overline6}{0,6}·0,\overline1 =$

$\small = \left[\left(0,\overline6\right)^{5+3}\right]^2÷\left(0,\overline6\right)^{2·7}-\dfrac{\dfrac{\cancel4^2}{\cancel{10}_5}}{\dfrac{4-0}{9}}·\dfrac{\dfrac{6-0}{9}}{\dfrac{\cancel6^3}{\cancel{10}_5}}·\dfrac{1-0}{9} =$

$\small = \left[\left(0,\overline6\right)^8\right]^2÷\left(0,\overline6\right)^{14}-\dfrac{\dfrac{2}{5}}{\dfrac{4}{9}}·\dfrac{\dfrac{\cancel6^2}{\cancel9_3}}{\dfrac{3}{5}}·\dfrac{1}{9} =$

$\small =\left(0,\overline6\right)^{8·2}÷\left(0,\overline6\right)^{14}-\dfrac{\cancel2^1}{5}·\dfrac{9}{\cancel4_2}·\dfrac{\dfrac{2}{3}}{\dfrac{3}{5}}·\dfrac{1}{9} =$

$\small =\left(0,\overline6\right)^{16}÷\left(0,\overline6\right)^{14}-\dfrac{1}{5}·\dfrac{9}{2}·\dfrac{2}{3}·\dfrac{3}{5}·\dfrac{1}{9} =$

$\small =\left(0,\overline6\right)^{16-14}-\dfrac{\cancel{90}^1}{\cancel{90}_1}·\dfrac{1}{9} =$

$\small =\left(0,\overline6\right)^2-\dfrac{1}{1}·\dfrac{1}{9} =$

$\small =\left(\dfrac{6-0}{9}\right)^2-\dfrac{1}{9} =$

$\small =\left(\dfrac{\cancel6^2}{\cancel9_3}\right)^2-\dfrac{1}{9} =$

$\small =\left(\dfrac{2}{3}\right)^2-\dfrac{1}{9} =$

$\small = \dfrac{4}{9}-\dfrac{1}{9} =$

$\small = \dfrac{\cancel3^1}{\cancel9_3} =$

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