@sabrina91 ...Ai presente come si avvolge ? Basta fare l'operazione inversa😉
30/100 = 3/10; si semplifica per 10;
(2/5 + 3/10) : (8/35 + 1/5) * 5/49 =
= (4/10 + 3/10) : (8/35 + 7/35) * 5/49 =
= 7/10 : 15/35) * 5/49; 15/35 = 3/7; si semplifica per 5;
= 7/10 : 3/7 * 5/49 =
= 7/10 * 7/3 * 5/49; 7 * 7 = 49; 49/49 = 1/1; 5/10 = 1/2; si semplifica per 5.
= 49 * 5 / (10 * 3 * 49) =
= 1/(2 * 3) = 1/6.
Ciao @sabrina91
(2/5 + 3/10) : (8 + 7)/35 * 5/49 =
= (4 + 3)/10 : 15/35 * 5/49 =
= 7/10 * 7/3 * 5/49 =
= 49/30 * 5/49 =
= 5/30 =
= 1/6
(2/5 + 3/10) : ((8 + 7)/35 * 5/49
7/10 : (15/35 * 5/49
7/10 * 35/15 * 5/49
7/10*7/3*5/49
7^2 e 49 si elidono
5/30
1/6
========================================================
$\small \left(\dfrac{2}{5}+\dfrac{\cancel{30}^3}{\cancel{100}_{10}}\right) : \left(\dfrac{8}{35}+\dfrac{1}{5}\right)·\dfrac{5}{49}=$
$\small = \left(\dfrac{2}{5}+\dfrac{3}{10}\right) : \left(\dfrac{8+7}{35}\right)·\dfrac{5}{49}=$
$\small = \left(\dfrac{4+3}{10}\right) : \left(\dfrac{\cancel{15}^3}{\cancel{35}_7}\right)·\dfrac{5}{49}=$
$\small = \dfrac{7}{10} : \dfrac{3}{7}·\dfrac{5}{49}=$
$\small = \dfrac{7}{10} · \dfrac{7}{3}·\dfrac{5}{49}=$
$\small = \dfrac{\cancel{49}^1}{\cancel{30}_6}·\dfrac{\cancel5^1}{\cancel{49}_1}=$
$\small = \dfrac{1}{6} · \dfrac{1}{1}=$
$\small = \dfrac{1}{6}$