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Alguien me podria ayudar con estos ejercicios

^ = elevado

a)(xy^1/2 – x^1/2+y^1/2)
*
(x^1/2+y1/2+x^-1/2 y)

b)(x^5/2-x^2+x^3/2-x+x^1/2+x^0)
*
(x^1/2+x^0)


c)3y^1/3(y^2/3-1)^2

2006-11-09 13:36:33 · 6 respuestas · pregunta de Dominic 1 en Ciencias y matemáticas Matemáticas

6 respuestas

a)(xy^1/2 – x^1/2+y^1/2) * (x^1/2+y^1/2+x^-1/2 y)

= x^1/2 xy^1/2 + y^1/2 xy^1/2 + x^-1/2 y xy^1/2 – x^1/2 x^1/2 – x^1/2 y^1/2 – x^1/2 x^-1/2 y + x^1/2 y^1/2 + y^1/2 y^1/2 + x^-1/2 y y^1/2
= x^3/2 y^1/2 + xy + x^1/2 y^3/2 – x – x^1/2 y^1/2 – y + x^1/2 y^1/2 + y + x^-1/2 y^3/2 (x^1/2 / x^-1/2)
= x^3/2 y^1/2 + xy + x^1/2 y^3/2 – x + x^1/2 y^3/2 / x
= x (x^1/2 y^1/2 + y - 1) + x^1/2 y^3/2 (1/x + 1)

= x (x^1/2 y^1/2 + y - 1) + x^1/2 y^3/2 (x^-1 + 1)

b)(x^5/2-x^2+x^3/2-x+x^1/2+x^0... * (x^1/2+x^0)

Todo elevado a la 0 es igual a 1
= (x^5/2 - x^2 + x^3/2 - x + x^1/2 + 1) * (x^1/2 + 1)
= x^5/2 x^1/2 - x^2 x^1/2 + x^3/2 x^1/2 - x x^1/2 + x^1/2 x^1/2 + x^1/2 + x^5/2 - x^2 + x^3/2 - x + x^1/2 + 1
= x^6/2 - x^2/2 + x^4/2 - x^3/2 + x^1/1 + x^1/2 + x^5/2 - x^2 + x^3/2 - x + x^1/2 + 1
= x^3 - x + x^2 - x^3/2 + x + x^1/2 + x^5/2 - x^2 + x^3/2 - x + x^1/2 + 1
= x^3 + x^5/2 - x + 2x^1/2 + 1

= x(x^2 - 1) + x^1/2(x^2 + 2) + 1

c)3y^1/3(y^2/3-1)^2

= 3y^1/3(y^4/3 - 2y^2/3 + 1)
= 3y^1/3 y^4/3 - 3y^1/3 2y^2/3 + 3y^1/3
= 3y^5/3 - 6y^3/3 + 3y^1/3

= 3y^5/3 - 6y + 3y^1/3

2006-11-12 01:03:30 · answer #1 · answered by atleyuquinnican 5 · 0 0

a)(xy¹/² – x¹/² + y¹/²)(x¹/² + y¹/² + x-¹/² y)=
x³/²y¹/² - x + x¹/²y¹/² + xy - x¹/²y¹/² + y + x³/²y³/² - xy + x¹/²y³/²


b)(x^5/2-x^2+x^3/2-x+x^1/2+x^0...) (x^1/2+x^0)
= x³ - x^5/2 + x² - x³/² + x + x¹/² + x^5/2 -x² + x³/² - x + x¹/²+ 1

c)3y¹/³(y²/³ - 1)² = 3y¹/³(y^4/3 - 2y²/³ + 1) = 3y^5/3 - 6y + 3y¹/³

'

2006-11-11 00:06:53 · answer #2 · answered by lola l 1 · 1 0

Mira Dominic se hacen de la siguiente manera los exponentes de la misma base se suman

a)(xy¹/² – x¹/² + y¹/²)
*
(x¹/² + y¹/² + x-¹/² y)=

x³/²y¹/² - x + x¹/²y¹/² + xy - x¹/²y¹/² + y + x³/²y³/² - xy + x¹/²y³/²


b)(x^5/2-x^2+x^3/2-x+x^1/2+x^0...
*
(x^1/2+x^0)

x³ - x^5/2 + x² - x³/² + x + x¹/² + x^5/2 -x² + x³/² - x + x¹/²+ 1

Ojo: x^0 = 1



c)3y¹/³(y²/³ - 1)²

3y¹/³(y^4/3 - 2y²/³ + 1)

3y^5/3 - 6y + 3y¹/³

Espero esto aclare tu duda

2006-11-09 23:51:01 · answer #3 · answered by ing_alex2000 7 · 1 0

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2006-11-09 22:11:29 · answer #4 · answered by mexicanita 3 · 0 1

se q el primero es raiz de x por raiz de y menos raiz de x + raiz y de y

(raiz cuadrada...)
pero hasta allii llega mi info (el cuadrado de 1/2 es la raiz de 2 )

2006-11-09 21:44:57 · answer #5 · answered by Luii 3 · 0 1

yo soy mala para las matematicas.

2006-11-09 21:41:01 · answer #6 · answered by Anonymous · 0 1

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