Question

Va rog frumos. Cine știe forare bn matematică ​

Answer 1

Răspuns:

Sper ca înțelegi din poza

Answer 2

adunarea și scăderea fracțiilor:

a)

$2\dfrac{1}{7} + 1\dfrac{3}{7} = (2+1) + \bigg(\dfrac{1}{7} + \dfrac{3}{7}\bigg) = 3 + \dfrac{1+3}{7} = \bf3\dfrac{4}{7}$

sau:

$2\dfrac{1}{7} + 1\dfrac{3}{7} = \dfrac{2 \cdot 7 + 1}{7} + \dfrac{1 \cdot 7 + 3}{7} = \dfrac{14 + 1}{7} + \dfrac{7 + 3}{7} = \dfrac{15}{7} + \dfrac{10}{7} = \dfrac{15+10}{7} = \dfrac{25}{7} = \dfrac{3 \cdot 7 + 4}{7} = \bf 3\dfrac{4}{7}$

b)

$1\dfrac{2}{5} + 3\dfrac{1}{10} = (1+3) + \bigg(\dfrac{^{2)} 2}{5} + \dfrac{1}{10}\bigg) = 4 + \dfrac{4+1}{10} = 4 + \dfrac{5^{(5} }{10} = \bf 4\dfrac{1}{2}$

sau:

$1\dfrac{2}{5} + 3\dfrac{1}{10} = \dfrac{1 \cdot 5 + 2}{5} + \dfrac{3 \cdot 10 + 1}{10} = \dfrac{5+2}{5} + \dfrac{30+1}{10} = \dfrac{^{2)} 7}{5} + \dfrac{31}{10} = \dfrac{14+31}{10} = \dfrac{45^{(5} }{10} = \dfrac{9}{2} = \dfrac{4 \cdot 2 + 1}{2} = \bf4\dfrac{1}{2}$

c)

$3\dfrac{1}{3} + 2\dfrac{2}{5} = (3+2) + \bigg(\dfrac{^{5)} 1}{3} + \dfrac{^{3)}2}{5}\bigg) = 5 + \dfrac{5+6}{15} = \bf 5\dfrac{11}{15}$

d)

$3\dfrac{5}{21} + 1\dfrac{7}{12} = (3+1) + \bigg(\dfrac{^{4)} 5}{21} + \dfrac{^{7)}7}{12}\bigg) = 4 + \dfrac{20+49}{84} = 4 + \dfrac{69^{(3} }{84} = \bf 4\dfrac{23}{28}$

e)

$5\dfrac{5}{6} - 1\dfrac{1}{6} = (5-1) + \bigg(\dfrac{5}{6} - \dfrac{1}{6}\bigg) = 4 + \dfrac{5-1}{6} = 4 + \dfrac{4^{(2} }{6} = \bf 4\dfrac{2}{3}$

f)

$4\dfrac{5}{18} - 3\dfrac{1}{9} = (4-3) + \bigg(\dfrac{5}{18} - \dfrac{^{2)} 1}{9}\bigg) = 1 + \dfrac{5-2}{18} = 1 + \dfrac{3^{(3} }{18} = \bf 1\dfrac{1}{6}$

g) (observ că $\frac{1}{7}$  este mai mic decât $\frac{2}{3}$ , am scădere ⇒ mă "împrumut" cu 1 întreg)

$7\dfrac{1}{7} - 4\dfrac{2}{3} = 6 + 1\dfrac{1}{7} - 4\dfrac{2}{3} = (6 - 4) + \bigg(1\dfrac{1}{7} - \dfrac{2}{3}\bigg) = 2 + \bigg(\dfrac{7 + 1}{7} - \dfrac{2}{3}\bigg)=$

$= 2 + \bigg(\dfrac{^{3)} 8}{7} - \dfrac{^{7)}2}{3}\bigg) = 2 + \dfrac{24-14}{21} = \bf 2 \dfrac{10}{21}$

sau

$7\dfrac{1}{7} - 4\dfrac{2}{3} = \dfrac{7 \cdot 7 + 1}{7} - \dfrac{4 \cdot 3 + 2}{3} = \dfrac{49 + 1}{7} - \dfrac{12 + 2}{3} = \dfrac{^{3)} 50}{7} - \dfrac{^{7)} 14}{3} = \dfrac{150}{21} - \dfrac{98}{21} = \dfrac{150 - 98}{21} = \dfrac{52}{21} = \dfrac{2 \cdot 21 + 10}{21} = \bf 2\dfrac{10}{21}$

h)

$9\dfrac{2}{3} - 8\dfrac{1}{4} = (9-8) + \bigg(\dfrac{^{4)}2}{3} - \dfrac{^{3)} 1}{4}\bigg) = 1 + \dfrac{8-3}{12} = \bf 1\dfrac{5}{12}$