utilizando solo operadores matematicos " + , - , * , / , ( , ) , sqtr, exp , x ala y , ...."
2 2 2 = 6
3 3 3 = 6
4 4 4 = 6
5 5 5 = 6
6 6 6 = 6
7 7 7 = 6
8 8 8 = 6
9 9 9 = 6
Suerte...
2006-12-07 07:30:58 · 11 respuestas · pregunta de Alfredo.Pineda 2 en Ciencias y matemáticas ➔ Matemáticas
2+2+2=6
3+(sqrt(3)*sqrt(3))=6
sqrt(4)+(sqrt(4)*sqrt(4))=6
5+(5/5)=6
6*(6/6)=6
7-(7/7)=6
8-sqrt(sqrt(8+8))=6
(sqrt(9)*sqrt(9))-sqrt(9)=6
2006-12-07 08:10:07 · answer #1 · answered by mexican 2 · 0⤊ 1⤋
2 + 2 + 2 = 6
3 + 3 ÷ 3° = 6
4 + 4 - √4 = 6
√(5 X 5) + 5° = 6
6 - 6 + 6 = 0
7 x 7° - 7° = 6
8 + raiz cúbica de 8 x 8° = 6
9 - √9 ÷ 9° = 6
2006-12-07 07:49:49 · answer #2 · answered by marcojarrin64 7 · 1⤊ 0⤋
2+2+2
3X3-3
5*5+5
6X6*6
QUE HUEVA
2006-12-07 07:36:06 · answer #3 · answered by Anonymous · 1⤊ 1⤋
2+2+2 =6
(3*3)-3 = 6
(sqrt de 4)+(sqrt de 4)+(sqrt 4) = 6
(5 / 5 ) + 5 = 6
-(7 / 7) + 7= 6
(raíz cúbica 8)+ (raíz cubica 8) + (raíz cubica 8) = 6
(sqrt (9*9)) - (sqrt 9) = 6
tené en cuenta q usen bien los signos y los paréntesis
2006-12-07 08:00:01 · answer #4 · answered by Anonymous · 0⤊ 1⤋
2 + 2 + 2 = 6
3 X 3 - 3 = 6
5 + 5 / 5 = 6
6 X 6 / 6 = 6
7 - 7 / 7 = 6
2 y 2 son 4, 4 y 2 son 6. listo!!!!.ademas, 6 y 2 son 8, y 8, 16 (Brinca la tablita yo la la brinque)
2006-12-07 07:47:01 · answer #5 · answered by BlutSauger 3 · 0⤊ 1⤋
2+2+2=6
(3*3)-3=6
SQRT4+SQRT4+SQRT4=6
(5/5)+5=6
(6*6)/6=6
((7 AL CUADRADO)-7)/7=6
8-SQRT(SQRT(8+8))=6
(9+9)/SQRT9=6
2006-12-07 07:45:14 · answer #6 · answered by SHALOM 4 · 0⤊ 1⤋
2+2+2= 6
3*3-3 = 6
sqrt4+sqrt4+sqrt4 = 6
5+(5/5)=6
7-(7/7)= 6
8 - sqrt(sqrt (8+8)) = 6
(9+9) / sqrt9 = 6
2006-12-07 07:42:43 · answer #7 · answered by silvia g 6 · 0⤊ 1⤋
2+2+2 6
3*3-3 6
2006-12-07 07:34:59 · answer #8 · answered by Anonymous · 0⤊ 1⤋
2 +2+ 2 = 6
3 *3- 3 = 6
4 4 4 = 6
(5/ 5)+ 5 = 6
6- 6 +6 = 6
7- (7/ 7) = 6
8 8 8 = 6
9 9 9 = 6
puff... starìa cabrón con los demás.. no soy un chico muy aplicado a las matemáticas... JAJAJA
2006-12-07 07:34:28 · answer #9 · answered by Efraín Herrera 4 · 0⤊ 1⤋
6=6
2006-12-07 07:34:32 · answer #10 · answered by Anonymous · 0⤊ 2⤋