Question

Simplificați fracția 2+5+8+...+362 supra 3+7+11+...+483,astfel încât să obții o fr ireductibila

Answer 1

2+5+8+...+362=
N=(362-2):3+1
N=360:3+1
N=120+1
N=121
S=(362+2)•121:2
S=364•121:2
S=44 044:2
S=22 022

3+7+11+...+483=?
N=(483-3):4+1
N=480:4+1
N=120+1
N=121
S=(483+3)•121:2
S=486•121:2
S=29 403
$\frac{22022}{29403}$
simplificam cu 11:
$\frac{2002}{2673}$
simplificam cu 11:
182/243

sper că te-am ajutat la ceva


Redeyes2

Answer 2

$\dfrac{2+5+8+...+362}{3+7+11+...+483} = \dfrac{(1\cdot 3-1)+(2\cdot 3-1)+...+(121\cdot 3-1)}{(1\cdot 4-1)+(2\cdot 4-1)+...+(121\cdot 4-1)} = \\ \\\\ = \dfrac{1\cdot 3+2\cdot 3+...+121\cdot 3-1-1-\overset{\text{de 121 ori}}{\overbrace{...}}-1}{1\cdot 4+2\cdot 4+...+121\cdot 4-1-1-\underset{\text{de 121 ori}}{\underbrace{...}}-1} = \\ \\\\ = \dfrac{3\cdot(1+2+...+121)-121}{4\cdot(1+2+...+121)-121} = \dfrac{3\cdot \dfrac{121\cdot 122}{2}-121}{4\cdot \dfrac{121\cdot 122}{2}-121} =$

$= \dfrac{3\cdot 121\cdot 61-121}{4\cdot 121\cdot 61-121} = \dfrac{121\cdot(3\cdot 61-1)}{121\cdot (4\cdot 66-1)} = \dfrac{3\cdot 61-1}{4\cdot 61-1} = \dfrac{182}{243}$