Question

Salut! Mă poate ajuta cineva și pe mine? Vă rog frumos! ☁☀​

Answer 1

EXERCIȚIUL 5:

a+b=49

(a,b)=7 => a=7x

b=7y, unde (x,y)=1

7x+7y=49

7(x+y)=49 |:7

x+y=7

x+y=7 }

(x,y)=1 } => (x,y)={(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)}

x,y sunt nr. nat. }

EXERCIȚIUL 6:

a) amplificare cu 2

$\frac{1}{3} = > \frac{2}{6} \\ \frac{4}{5} = > \frac{8}{10} \\ \frac{7}{8} = > \frac{14}{16} \\ \frac{5}{6} = > \frac{10}{12} \\ \frac{2x}{3y} = > \frac{4x}{6y} \\ \frac{a + 1}{a + 2} = > \frac{2a + 2}{2a + 4}$

b) amplificare cu 3

$\frac{1}{3} = > \frac{3}{9} \\ \frac{4}{5} = > \frac{12}{15} \\ \frac{7}{8} = > \frac{21}{24} \\ \frac{5}{6} = > \frac{15}{18} \\ \frac{2x}{3y} = > \frac{6x}{9y} \\ \frac{a + 1}{a + 2} = > \frac{3a + 3}{3a + 6}$

c) amplificare cu 5

$\frac{1}{3} = > \frac{5}{15} \\ \frac{4}{5} = > \frac{20}{25} \\ \frac{7}{8} = > \frac{35}{40} \\ \frac{5}{6} = > \frac{25}{30} \\ \frac{2x}{3y} = > \frac{10x}{15y} \\ \frac{a + 1}{a + 2} = > \frac{5a + 5}{5a + 10}$

d)k, k aparține N*

$\frac{1}{3} = > \frac{k}{3k} \\ \frac{4}{5} = > \frac{4k}{5k} \\ \frac{7}{8} = > \frac{7k}{8k} \\ \frac{5}{6} = > \frac{5k}{6k} \\ \frac{2x}{3y} = > \frac{2xk}{3yk} \\ \frac{a + 1}{a + 2} = > \frac{k(a + 1)}{k(a + 2)} = \frac{ak + k}{ak + 2k}$