CLASE 24
Calcula aplicando las propiedades de los radicales. 2 + 22 22 22 22 22 22 3 3 22 22 + + 22 22 + + b) a) 66 66 4 4 55 55 4 4 33 33 4 4
2 + 22 22 22 22 22 22 3 3 22 22 + + 22 22 + + a) 2 + 22 22 22 22 2222 2222 2 – 22 22 ( ( ) ) ( ( ) ) 22 22 ( ( ) ) = = – – 22 22 ( ( ) ) – – 22 22 ( ( ) ) = = 2222 2222 – – – 2 4 – 2 = = 2 2 2 – – 1 1 ( ( ) ) 2 2 = = – – 1 1 – – 1 1 = = 22 22 22 22 3 3 22 22 2 2 2222 2222 = = 3 3 88 8 = = = = 22 22 + + 22 2 2 – – 1 1 = =
Halla el valor de A en cada caso. 4x4x 4x4x b) a) ( ( x x ) ) 3 3 A A 5 5 = = (x > 0) ( 10 – 5 ) 15 – A 2 2 = =
4x4x 4x4x a) ( ( x x ) ) 3 3 A A 5 5 = = ) ) ( ( x x x x ( ( ) ) ( ( ) ) 3 3 = = x x = = 3 3 = = x x
4x4x 4x4x a) ( ( x x ) ) 3 3 A A 5 5 = = 3 3 x x 3 3 x x 4x4x 4x4x A A 5 5 = = 3 3 x x A A = = 4x4x 4 x 3 3 A A = = 4 4
4x4x 4x4x a) ( ( x x ) ) 3 3 A A 5 5 = = 3 3 x x 4x4x 4x4x A A 5 5 = = 3 3 x x A A = = 4x4x 4x4x 3 3 A A = = A A = = A A = = = 16 = 16
Una de las soluciones de la ecuación x 2 – 4 x – 1 = 0 es: 5 5 5 5 2 + 5 a) 2 5 b) 5 – 2 c) d)
El valor de es: 2 3 + 11 2 3 – 11 a) 27 b) 1 1 c) d)