Răspuns:
9.a)
[tex]\frac{1}{30*31}+\frac{1}{31*32}+\frac{1}{32*33}+...+\frac{1}{89*90}=\frac{31-30}{30*31}+\frac{32-31}{31*32}+\frac{33-32}{32*33}+...+\frac{90-89}{89*90}=\frac{31}{30*31}-\frac{30}{30*31}+\frac{32}{31*32}-\frac{31}{31*32}+\frac{33}{32*33}-\frac{32}{32*33}+...+\frac{90}{89*90}-\frac{89}{89*90}=\frac{1}{30}-\frac{1}{31}+\frac{1}{31}-\frac{1}{32}+\frac{1}{32}-\frac{1}{33}+...+\frac{1}{89}-\frac{1}{90}=\frac{1}{30}-\frac{1}{90}=\frac{3-1}{90}=\frac{2}{90}=\frac{1}{45}[/tex]
b)
[tex]\frac{3}{5*8}+\frac{3}{8*11}+\frac{3}{11*14}+...+\frac{3}{47*50}=\frac{8-5}{5*8}+\frac{11-8}{8*11}+\frac{14-11}{11*14}+...+\frac{50-47}{47*50}=\frac{8}{5*8}-\frac{5}{5*8}+\frac{11}{11*8}-\frac{8}{11*8}+\frac{14}{11*14}-\frac{11}{14*11}+...+\frac{50}{50*47}-\frac{47}{47*50}=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{11}{47}-\frac{1}{50}=\frac[1}{5}-\frac{1}{50}=\frac{10-1}{50}=\frac{9}{50}[/tex]
10. [tex]\frac{15}{2x+1}[/tex]∈N ⇔2x+1∈D15
D15={1,3,5,15}
2x+1=1 2x=0 x=0
2x+1=3 2x=2 x=1
2x+1=5 2x=4 x=2
2x+1=15 2x=14 x=7
Explicație pas cu pas:
