Productos notables 1. Binomio al cuadrado ο Suma al cuadrado (π + π)2 = π β2 + 2ππ + π 2 β πππππππ ππ ππ’ππππππ π‘ππππππ
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Productos notables
1. Binomio al cuadrado ο Suma al cuadrado (π + π)2 = π β2 + 2ππ + π 2 β πππππππ ππ ππ’ππππππ
π‘πππππππ ππ’ππππππ πππππππ‘π
Es decir el desarrollo de (π + π)2es: Cuadrado del primero π2 MΓ‘s el doble del primero por el segundo +2ππ Mas el cuadrado del segundo +π 2 ο Resta al cuadrado (π β π)2 = π β2 β 2ππ + π 2 β πππππππ ππ ππ’ππππππ
π‘πππππππ ππ’ππππππ πππππππ‘π
Es decir el desarrollo de (π + π)2es: Cuadrado del primero π2 Menos el doble del primero por el segundo β2ππ Mas el cuadrado del segundo +π 2 Nota: (π β π)2 = (π β π)2 AdemΓ‘s si π2 = 0 πππ‘πππππ π = 0 (π β π)2 = 0 πππ‘πππππ π β π = 0 Demostracion (π + π)2 = π2 + 2ππ + π 2 (π + π)2 = (π + π)(π + π) por la ley distributiva (π + π)π = ππ + ππ =(β π+β π) β (π + π) = π(π + π) + π(π + π) por la ley distributiva = π. π + π. π + π. π + π. π Por la ley conmutativa ππ = ππ = π2 + ππ + ππ + π 2 Reduciendo tΓ©rminos semejantes (π + π)2 = π2 + 2ππ + π 2 DemostraciΓ³n geometrica: El area del cuadrado grande es igual al area de sus partes
π
π
π
ππ
π2
π
π
π2
ππ
π =
βπ
π (π+π)2
π2
+
ππ
+
= β π2 +2ππ+π2
ππ
+
π2
Ejemplos ο (π + 2)2 = (π)2 + 2(π)(2) + (2)2 (π + 2)2 = π2 + 4π + 4 ο (2π₯ + 3π¦)2 = (2π₯)2 + 2(2π₯)(3π¦) + (3π¦)2 (2π₯ + 3π¦)2 = 4π₯ 2 + 12π₯π¦ + 9π¦ 2 ο (π₯ π + π¦ π )2 = (π₯ π )2 + 2(π₯ π )(π¦ π ) + (π¦ π )2 (π₯ π + π¦ π )2 = π₯ 2π + 2π₯ π π¦ π + π¦ 2π ο (π β 3)2 = (π)2 β 2(π)(3) + (3)2 (π β 3)2 = π2 β 6π + 9 ο (3π₯ β 2π¦)2 = (3π₯)2 β 2(3π₯)(2π¦) + (2π¦)2 (3π₯ β 2π¦)2 = 9π₯ 2 β 12π₯π¦ + 4π¦ 2 ο (π₯ π β π¦ π )2 = (π₯ π )2 β 2(π₯ π )(π¦ π ) + (π¦ π )2 (π₯ π β π¦ π )2 = π₯ 2π β 2π₯ π π¦ π + π¦ 2π 1 2
1
1 2
ο (π₯ + π₯) = (π₯)2 + 2(π₯) (π₯) + (π₯) 1 2 1 (π₯ + ) = π₯ 2 + 2 + 2 π₯ π₯
El proceso inverso es llevar de un trinomio cuadrado perfecto a un binomio al cuadrado ο π₯ 2 + 2π₯ + 1 = (π₯)2 + 2(π₯)(1) + (1)2 = (π₯ + 1)2 ο π₯ 2 β 2π₯ + 1 = (π₯)2 β 2(π₯)(1) + (1)2 = (π₯ β 1)2 2 ο π₯ β 6π₯ + 9 = (π₯)2 β 2(π₯)(3) + (3)2 = (π₯ β 3)2
2. Identidades de legendre Son identidades que surgen a partir del binomio al cuadrado (π + π)2 + (π β π)2 = 2(π2 + π 2 )
...(I)
(π + π)2 β (π β π)2 = 4ππ
β¦(II)
(π + π)4 β (π β π)4 = 8ππ(π2 + π 2 ) β¦(III)
DemostraciΓ³n de (I) (π + π)2 + (π β π)2 = 2(π2 + π 2 ) Desarrollando cada binomio al cuadrado (π π2 + 2ππ + π 2 + β π2 β 2ππ + π 2 β + π)2 + (π β β π)2 = β Reduciendo tΓ©rminos semejantes (π + π)2 + (π β π)2 = 2π2 + 2π 2 Factorizando el 2
(π + π)2 + (π β π)2 = 2(π2 + π 2 ) Demostracion de (II) (π + π)2 β (π β π)2 = 4ππ Desarrollando cada binomio al cuadrado (π π2 + 2ππ + π 2 β (π β + π)2 β (π β β π)2 = β β 2 β 2ππ + π 2 ) = π2 + 2ππ + π 2 β π2 + 2ππ β π 2 Reduciendo tΓ©rminos semejantes (π + π)2 + (π β π)2 = 4ππ DemostraciΓ³n de (III) (π + π)4 β (π β π)4 = 8ππ(π2 + π 2 Aplicando teoria de exponentes π₯ 4 = (π₯ 2 )2 ley de potencia de potencia 2
(π + π)4 β (π β π)4 = [(π β + π)2 ] β [(π β β π)2 ]
2
Aplicando el desarrollo de binomio al cuadrado 2
2
(π + π)4 β (π β π)4 = [π β2 + 2ππ + π 2 ] β [π β2 β 2ππ + π 2 ] Agrupando de manera adecuada = [(π2 + π 2 ) + 2ππ]2 β [(π2 + π 2 ) β 2ππ]2 Haciendo el cambio de variable (π2 + π 2 ) = π y 2ππ = π pueda observarse mejor 2
2
2
2
2
2
= [(π β ] β [(π β] β + π ) + 2ππ β + π ) β 2ππ π
π
π
π
= [π + π]2 β [π β π]2 Podemos aplicar la segunda identidad de legendre = 4ππ retornando a los valores originales π = (π2 + π 2 ) y π = 2ππ = 4(π2 + π 2 ) (2ππ) 4 4 (π + π) β (π β π) = 8ππ(π2 + π 2 ) Tambien podemos aplicar diferencia de cuadrados en la demostraciΓ³n.
3. Diferencia de cuadrados (π + π)(π β π) =
β2 β π 2 π ππππππππππ ππ ππ’πππππππ
DemostraciΓ³n (π + π)(π β π) = (π + π)(π β π) = (π + π)(π β π) = π. π β π. π + π. π β π. π Por la ley conmutativa ππ = ππ = π2 β ππ + ππ β π 2 Reduciendo tΓ©rminos semejantes (π + π)(π β π) = π2 β π 2 Ejemplos ο (π + π)(π β π) = π2 β π2
ο (π₯ + π)(π₯ β π) = π₯ 2 β π2 ο (π₯ β 1)(π₯ + 1) = (π₯)2 β (1)2 = π₯2 β 1 ο (π₯ β 2)(π₯ + 2) = (π₯)2 β (2)2 = π₯2 β 4 2 2 ο (π β +β π 2 )(π β ββ π 2 ) = (π β2 )2 β (π β2 )2 π
π
π
π
π
= π4 β π4
π
ο (π20 + π40 )(π20 β π40 ) = (π20 )2 β (π40 )2 = π40 β π80 ο ( 2π₯ β + 3π¦ β β 3π¦ β )2 β (3π¦ β )(2π₯ β ) = (2π₯ β )2 π
π
π
π
π
= 4π₯ 2 β 9π¦ 2
π
ο (5π₯ β 7π¦)(5π₯ + 7π¦) = (5π₯)2 β (7π¦)2 = 25π₯ 2 β 49π¦ 2 ο (ππ + π π )(ππ β π π ) = (ππ )2 β (π π )2 = π2π β π 2π 2
2
ο (βπ₯ + βπ¦)(βπ₯ β βπ¦) = (βπ₯) β (βπ¦) =π₯βπ¦
ο (ππ+1 + π π+3 )(ππ+1 β π π+3 ) = (ππ+1 )2 β (π π+3 )2 = π(π+1)2 β π (π+3)2 = π2π+2 β π 2π+6 El proceso inverso de una diferencia de cuadrados llevarlo a un producto Simplemente sacamos la mitad de cada exponente ο π2 β π2 = (π + π)(π β π) ο π4 β π2 = (π2 + π)(π2 β π) ο π8 β π6 = (π4 + π3 )(π4 β π3 ) ο π40π β π10π = (π20π + π5π )(π20π β π5π ) ο π2π β π 2π = (ππ + π π )(ππ β π π ) ο (2π₯)2 β (3π¦)2 = (2π₯ + 3π¦)(2π₯ β 3π¦)
3. Binomio al cubo (π + π)3 = π3 + 3π2 π + 3ππ 2 + π 3 (π + π)3 = β π3 + π 3 + 3ππ(π + π) Forma corta πππππ‘ππππ ππ πΆππ’πβπ¦
(π β π)3 = π3 β 3π2 π + 3ππ 2 β π 3 (π β π)3 = β π3 β π 3 β 3ππ(π β π) Forma corta πππππ‘ππππ ππ πΆππ’πβπ¦
DemostraciΓ³n Demostraremos el primero los demΓ‘s son anΓ‘logos (π + π)3 = β (π + π)(π + π) (π + π) =β (π + π)2 (π + π) Aplicando binomio al cuadrado =β (π2 + 2ππ + π 2 ) (π + π) Por la ley distributiva = (π β 2 + 2ππ + π 2 ) π + (π β 2 + 2ππ + π 2 ) π Otra vez la ley distributiva =β π3 + 2π2 π + π 2 π + β π2 π + 2ππ 2 + π 3 Propiedad conmutativa π 2 π = ππ 2 = π3 + 2π2 π + ππ 2 + π2 π + 2ππ 2 + π 3 Reduciendo terminos semejantes (π + π)3 = π3 + 3π2 π + 3ππ 2 + π 3 Ejemplos Aplicando la forma larga (π + π)3 = π3 + 3π2 π + 3ππ 2 + π 3 ο (2π₯ β + 3π¦ β )3 = (2π₯)3 + 3(2π₯)2 (3π¦) + 3(2π₯)(3π¦)2 + (3π¦)3 π
π
= 8π₯ 3 + 3(4π₯ 2 )(3π¦) + 3(2π₯)(9π¦ 2 ) + 27π¦ 3 = 8π₯ 3 + 36π₯ 2 π¦ + 54π₯π¦ 2 + 27π¦ 3
ο (π₯ + 2π¦)3 = (π₯)3 + 3(π₯)2 (2π¦) + 3(π₯)(2π¦)2 + (2π¦)3 = π₯ 3 + 6π₯ 2 π¦ + 3π₯(4π¦ 2 ) + 8π¦ 3 = π₯ 3 + 6π₯ 2 π¦ + 12π¦ 2 + 8π¦ 3 1 3
ο (π₯ + π₯)
1
1
= (π₯)3 + 3(π₯)2 (π₯) + 3(π₯)(π₯)2 + (π₯)3 1
1
1
= π₯ 3 + 3π₯ 2 . π₯ + 3π₯. π₯ + (π₯)3 = π₯ 3 + 3π₯ + 3
1 π₯
1
+ π₯3
ο (2π β β 3π β )3 = (2π β )2 (3π β ) + 3(2π β )( β 3π )2 β (3π β )3 β )3 β 3(2π π
π
π
π
π
π
π
= 8π3 β 3(4π2 )(3π) + 3(2π)(9π2 ) β 27π3 = π3 β 36π2 π + 54ππ2 β 27π3 Aplicando la forma corta (π + π)3 = π3 + π 3 + 3ππ(π + π) (π β π)3 = π3 β π 3 β 3ππ(π β π) 1 3 π₯
1 π₯
1 π₯
1 π₯
ο (π₯ + ) = (π₯)3 + ( ) 3 + 3(π₯)( )(π₯ + ) 1
1
= π₯ 3 + π₯ 3 + 3(π₯ + π₯)
π
ο (2π₯ + 1)3 = (2π₯)3 + (1)3 + 3(2π₯)(1)(2π₯ + 1) = 8π₯ 3 + 1 + 6π₯(2π₯ + 1) ο (π β π)3 = (π)3 β (π)3 β 3(π)(π)(π β π) ο (2π₯ β 3)3 = (2π₯)3 β (3)3 β 3(2π₯)(3)(2π₯ β 1) = 8π₯ 3 β 27 β 18π₯(2π₯ β 3)
4. Suma de cubos y Diferencia de cubos π3 + π 3 = (π + π)(π2 β ππ + π 2 ) π3 β π 3 = (π β π)(π2 + ππ + π 2 ) Demostracion Solo demostraremos la p rimera la siguiente es anΓ‘logo (π + π)(π2 β ππ + π 2 ) = (π + π) (π β 2 β ππ + π 2 ) π
Por la ley distributiva (π + π)π = ππ + ππ = π(π2 β ππ + π 2 ) + π(π2 β ππ + π 2 ) Por la ley distributiva =β π. π2 β π. ππ + π. π 2 + β π. π2 β π. ππ + π. π 2 = π3 β π2 π + ππ 2 + π2 π β ππ 2 + π 3 Reduciendo tΓ©rminos semejantes (π + π)(π2 β ππ + π 2 ) = π3 + π 3
5. Identidades de Stevin (π₯ + π)(π₯ + π) = π₯ β2 + (π β + π)π₯ + π₯.π₯
π π’ππ
ππ β πππππ’ππ‘π
(π₯ + π)(π₯ + π)(π₯ + π) = π₯ 3 + (π + π + π)π₯ 2 + (ππ + ππ + ππ)π₯ + πππ DemostraciΓ³n (π₯ + π)(π₯ + π) = (π₯ + π) β (π₯ + π) Por la ley distributiva (π₯ + π)(π₯ + π) = π₯ β (π₯ + π) + π β (π₯ + π) Por la ley distributiva (π₯ + π)(π₯ + π) = π₯. π₯ + π₯. π + π. π₯ + π. π ordenando (π₯ + π)(π₯ + π) = π₯ 2 + ππ₯ + ππ₯ + ππ Factorizando la x (π₯ + π)(π₯ + π) = π₯ 2 + (π + π)π₯ + ππ Ejemplos ο Efectuar (π₯ β 5)(π₯ + 7) ππ’ππ β 5 + 7 = +2 ; πππππ’ππ‘π = (β5)(+7) = β35 (π₯ β 5)(π₯ + 7) = π₯ 2 + 2π₯ β 35 ο Efectuar (π₯ β 4)(π₯ β 3) ππ’ππ = β4 β 3 = β7 πππππ’ππ‘π = (β4)(β3) = +12 2 (π₯ β 2)(π₯ β 3) = π₯ β 7π₯ + 12
ο Efectuar (π + 2)(π + 10) ππ’ππ = +2 + 10 = +12 πππππ’ππ‘π = (+2)(+10) = +20 (π + 2)(π + 10) = π2 + 12π + 20 ο (π β 5)(π β 6) = π2 β 11π + 30 ο Efectuar (π β2 + 2)(π β2 + 10) = (π β2 )2 + 12 π β2 + 20 π₯
π₯
π₯
π₯
(π2 + 2)(π2 + 10) = π4 + 12π2 + 20 ο Efectuar (π3 + 2)(π3 + 3) = (π β3 )2 + 5 π β3 + 6 π₯
π₯
= π6 + 5π3 + 6 π₯ π₯ ο Efectuar (π + 5)(π + 3) = (π π₯ )2 + 8π π₯ + 15 = π2π₯ + 8π π₯ + 15 ο Efectuar (π π₯+1 + 2)(π π₯+1 + 3) = (π π₯+1 )2 + 5π π₯+1 + 6 = π(π₯+1)2 + 5π π₯+1 + 6 = π2π₯+2 + 5π π₯+1 + 6 2 3 2 3 ο Efectuar (ππ π + 4)(ππ π + 3) = (ππ 2 π 3 )2 + 7ππ 2 π 3 + 12 = π1.2 π2.2 π 3.2 + 7ππ 2 π 3 + 12 = π2 π 4 π 6 + 7ππ 2 π 3 + 12
6. Trinomio al cuadrado
(π + π + π)2 = π2 +π 2 +π 2 + 2ππ + 2ππ + 2ππ
DemostraciΓ³n (π + π + π)2 = (π + π + π)(π β+ π + π ) π
πππ ππ πππ¦ πππ π‘ππππ’π‘ππ£π (π + π + π)π = π. π + π. π + π. π (π + π + π)2 = π. (π β+ π + π ) + π. (π β+ π + π ) + π. (π β+ π + π ) Por la ley distributiva otra vez = π2 + ππ + ππ + ππ + π 2 + ππ + ππ + ππ + π 2 πππ ππ πππ¦ πππππ’π‘ππ‘ππ£π ππ = ππ π¦ ππ = ππ = π2 + ππ + ππ + ππ + π 2 + ππ + ππ + ππ + π 2 Reduciendo tΓ©rminos semejantes (π + π + π)2 = π2 +π 2 +π 2 + 2ππ + 2ππ + 2ππ
7. Trinomio al cubo
(π + π + π)3 = π3 + π 3 + π 3 + 3π2 π + 3π2 π + 3π 2 π + 3π 2 π + 3π 2 π + 3π 2 π + 6πππ (π + π + π)3 = π3 + π 3 + π 3 + 3(π + π)(π + π)(π + π)Forma corta
DemostraciΓ³n (π + π + π)3 = (π + π + π)(π + π + π)(π + π + π) =β (π + π + π)2 (π + π + π) Aplicando trinomio al cuadrado =β (π2 +π2 +π 2 + 2ππ + 2ππ + 2ππ) (π + π + π) Por la ley distributiva = (π2 +π 2 +π 2 + 2ππ + 2ππ + 2ππ)π +(π2 +π2 +π 2 + 2ππ + 2ππ + 2ππ)π +(π2 +π2 +π 2 + 2ππ + 2ππ + 2ππ)π Otra vez aplicando la ley distributiva
= π3 +ππ2 +ππ 2 + 2π2 π + 2π2 π + 2πππ +π2 π + π 3 +ππ 2 + 2ππ 2 + 2πππ + 2π 2 π +π2 π+π 2 π + π 3 + 2πππ + 2ππ 2 + 2ππ 2 Reduciendo terminos semejantes (π + π + π)3 = π3 + π 3 + π 3 + 3π2 π + 3π2 π + 3π 2 π + 3π 2 π + 3π 2 π + 3π 2 π + 6πππ
8. Identidad de Argand (π₯ 2 + π₯π¦ + π¦ 2 )(π₯ 2 β π₯π¦ + π¦ 2 ) = (π₯ 4 + π₯ 2 π¦ 2 + π¦ 4 ) (π₯ 2 + π₯ + 1)(π₯ 2 β π₯ + 1) = (π₯ 4 + π₯ 2 + 1) Ejemplos
9. Identidad de lagrange
(ππ₯ + ππ¦)2 + (ππ₯ β ππ¦)2 = (π2 + π 2 )(π₯ 2 + π¦ 2 )
10. Identidades de gauss
π₯ 3 + π¦ 3 + π§ 3 β 3π₯π¦π§ = (π₯ + π¦ + π§)(π₯ 2 + π¦ 2 + π§ 2 β π₯π¦ β π₯π§ β π¦π§) (π₯ + π¦)(π₯ + π§)(π¦ + π§) + π₯π¦π§ = (π₯ + π¦ + π§)(π₯π¦ + π₯π§ + π¦π§)
11. Igualdades condicionales Si a+b+c=0 entonces se verifican: π2 + π 2 + π 2 = β2(ππ + ππ + ππ) π3 + π 3 + π 3 = 3πππ PROBLEMAS BASICOS 1. Si π + π = β5 π¦ ππ = 3 Calcular: (π β π)2 SoluciΓ³n Aplicano la identidad de Legendre (π + π)2 β (π β π)2 = 4ππ Reemplazando datos π + π = β5 π¦ ππ = 3 2
( β5 ) β (π β π)2 = 4(3) operando 5 β (π β π)2 = 12 β(π β π)2 = 12 β 5 β(π β π)2 = 7 (π β π)2 = β7 2. Sabiendo que π + π = 11; ππ = 20 Calcular: πΈ = βπ2 + π 2 SoluciΓ³n Elevando al cuadrado ambos miembros (π + π)2 = (11)2 Aplicando binomio al cuadrado (π + π)2 = π2 + 2ππ + π 2 π2 + 2ππ + π 2 =121 Por dato ab=20 π2 + 2(20) + π 2 = 121 π2 + 40 + π 2 = 121 π2 + π 2 = 121 β 40
π2 + π 2 = 81 πΈ = βπ2 + π 2 = β81 = 9 1 π₯
3. Si π₯ + = 2 1
Hallar:πΈ = π₯ 40 + π₯ 40 SoluciΓ³n 1 π₯+ =2 π₯ π₯2 + 1 =2 π₯ operando π₯ 2 + 1 = 2π₯ β2 β 2π₯ + 1 π₯
=0
π‘πππππππ ππ’ππππππ πππππππ‘π 2
(π₯ ββ 1) = 0 π₯ β 1 = 0; π₯ = 1 1 1 πΈ = (1)40 + =1+ =1+1=2 40 (1) 1 4. si (π₯ + π¦)2 = 2(π₯ 2 + π¦ 2 ) 3π₯+2π¦ Calcular: πΈ = 5π₯ SoluciΓ³n (π₯ + π¦)2 = 2(π₯ 2 + π¦ 2 ) β Aplicamos binomio al cuadrado(π + π)2 = π2 + 2ππ + π 2 π₯ β2 + 2π₯π¦ + π¦ 2 = 2π₯ 2 + 2π¦ 2 ππππ’πππππ π‘πππππππ π πππππππ‘ππ 2π₯π¦ = 2π₯ 2 + 2π¦ 2 βπ₯ 2 β π¦ 2 2π₯π¦ = π₯ 2 + π¦ 2 Pasamos a restar 0= β π₯ 2 β 2π₯π¦ + π¦ 2 π‘ππππππ ππ’ππππππ πππππππ‘π 2
0 = (π₯ ββ π¦) 0
π₯ β π¦ = 0; π₯ = π¦ reemplazamos 3π₯ + 2π₯ 5π₯ πΈ= = =1 5π₯ 5π₯ 5. Si a+b=5 π2 + π 2 = 21 Calcular: π3 + π 3 Solucion Aplicando binomio al cuadrado (π + π)2 = π2 + 2ππ + π 2 (π β+ π)2 = β π2 + π 2 + 2ππ Reemplazando datos (5)2 = 21 + 2ππ 2 = ππ
Aplicando binomio al cubo (π β+ π)3 = π3 + π 3 + 3 ππ β (π β+ π) 3 3 3 (5) = π + π + 3(2)(5) 95 = π3 + π 3 6. Si:
π2 π
β
π2 π
= 3(π β π) 3(π 4 +π4 ) π 2 π2
Calcular: πΈ = SoluciΓ³n π2 π 2 β = 3(π β π) π π Dando comΓΊn denominador π3 β π 3 = 3(π β π) ππ operando π3 β π 3 = 3ππ(π β π) π3 β π 3 β 3ππ(π β π) = 0 Es un binomio al cubo forma abreviada (π β π)3 = 0 πβπ =0 ; π =π reemplazando πΈ= πΈ=
3(π 4 +π4 ) π2 π2
=
3(π4 +π4 ) π2 π2
3(2π 4 ) 6π 4 = 4 =6 π2π2 π
7. Efectuar π = (π β 3)(π + 1) β (π + 3)(π β 5) soluciΓ³n π = (π β β 3)(π + 1) β (π β + 3)(π β 5) Aplicando stevin (π₯ + π)(π₯ + π) = π₯ 2 + (π + π)π₯ + ππ π = (π2 β 2π β 3) β (π2 β 2π β 15) π = π2 β 2π β 3 β π2 + 2π + 15) π = 12 8. Si π₯ 2 + 5π₯ + 3 = 0 Hallar: π = (π₯ + 1)(π₯ + 2)(π₯ + 3)(π₯ + 4) + 6 Solucion Ordenando el producto (π₯ + 1)(π₯ + 4) β (π₯ + 2)(π₯ + 3) + 6 π=β Aplicando stevin (π₯ + π)(π₯ + π) = π₯ 2 + (π + π)π₯ + ππ πΎ = (π₯ 2 + 5π₯ + 4)(π₯ 2 + 5π₯ + 6) + 6 Pero por dato π₯ 2 + 5π₯ + 3 = 0 Es decir π₯ 2 + 5π₯ = β3 reemplazando en K πΎ = (π₯ β2 + 5π₯ + 4) (π₯ β2 + 5π₯ + 6) + 6 πΎ = (β3 + 4)(β3 + 6) + 6 = (1)(+3) + 6 = 9 1
9. Si: π₯ + π₯ = 3
1
Calcular: π₯ 2 + π₯ 2
SoluciΓ³n Elevando al cuadrado ambos miembros 1 2 (π₯ + ) = (3)2 π₯ Aplicando binomio al cuadrado (π + π)2 = π2 + 2ππ + π 2 1 1 2 π₯ 2 + 2(π₯) ( ) + ( ) = 9 π₯ π₯ 1 2 π₯ +2+ 2 = 9 π₯ 1 π₯2 + 2 = 9 β 2 π₯ π₯2 +
1 =7 π₯2 1
10. Si: π₯ + π₯ = β5 1
Calcular: π₯ 4 + π₯ 4
SoluciΓ³n Elevando al cuadrado ambos miembros 1 2 2 (π₯ + ) = (β5) π₯ Aplicando binomio al cuadrado (π + π)2 = π2 + 2ππ + π 2 1 1 2 (π₯)2 + 2(π₯) ( ) + ( ) = 5 π₯ π₯ 1 2 π₯ +2+ 2 = 5 π₯ 1 π₯2 + 2 = 5 β 2 π₯ 1 2 π₯ + 2=3 π₯ Elevando otra vez al cuadrado 1 2 (π₯ 2 + 2 ) = (3)2 π₯ 1 1 2 (π₯ 2 )2 + 2(π₯ 2 ) ( 2 ) + ( 2 ) = 9 π₯ π₯ 1 4 π₯ +2+ 4 = 9 π₯ 1 π₯ 4 + π₯4 = 7 11. Reducir: π = (π + 4)3 β (π + 3)(π + 4)(π + 5) SoluciΓ³n Paraque sea mas fΓ‘cil se hace cambio de variable π + 4 = π π = (π β + 4)3 β (π β + 3)(π β + 4)(π β + 5) π
πβ1
π = π3 β (π β 1)(π)(π + 1) Ordenamos
π
π+1
π = π3 β β (π β 1)(π + 1) π ππ π’ππ ππππππππππ ππ ππ’πππππππ
Aplicando diferencia de cuadrados (π + π)(π β π) = π2 β π 2 π = π3 β (π2 β 1)π Operando π = π3 β (π3 β π) π = π3 β π3 + π π=π Volviendo a nuestra variable inicial a=m+4 π =π+4 12. Reducir: πΈ = (π + π)(π β π)(π2 + π 2 ) + (π 4 + π4 ) SoluciΓ³n (π + π)(π β π) (π2 + π 2 ) + (π 4 + π4 ) πΈ=β Aplicando diferencia de cuadrados (π + π)(π β π) = π2 β π 2 πΈ = (π2 β π 2 )(π2 + π 2 ) + (π 4 + π4 ) Aplicando diferencia de cuadrados (π2 β π 2 )(π2 + π 2 ) = (π2 )2 β (π 2 )2 = π4 β π 4 πΈ = π4 β π 4 + (π 4 + π4 ) πΈ = π4 β π 4 + π 4 + π4 πΈ = 2π4 1
13. Si π₯ + π₯ = 3 Calcular: πΈ = π₯3 + π₯2 +
1 1 + 2 3 π₯ π₯
SoluciΓ³n Observamos y ordenamos la suma 1 1 πΈ = π₯2 + 2 + π₯3 + 3 β π₯ β π₯ Calculando la suma de cuadrados Elevando al cuadrado ambos miembros 1 2 (π₯ + ) = (3)2 π₯ Aplicando binomio al cuadrado (π + π)2 = π2 + 2ππ + π 2 1 1 2 (π₯)2 + 2(π₯) ( ) + ( ) = 9 π₯ π₯ 1 =9 π₯2 1 π₯ 2 + π₯ 2 = 7β¦β¦β¦β¦β¦β¦(1) Calculando la suma de cubos Elevando al cubo ambos miembros 1 3 (π₯ + ) = (3)3 π₯ Aplicando binomio al cubo (π + π)3 = π3 + π 3 + 3ππ(π + π) π₯2 + 2 +
1 3
1
1
(π₯)3 + ( ) + 3(π₯)( ) (π₯ + ) =27 π₯ π₯ π₯ 1 3 π + 3 + 3. (3) = 27 π 1 3 π₯ + 3 = 18 β¦ β¦ β¦ . (2) π Reemplazando (1) y (2)en E πΈ = 7 + 18 = 27 14. si: π₯ 2 + π¦ 2 = 8 π₯+π¦ =4 π₯ π¦ Hallar: π = π¦ + π₯ SoluciΓ³n Trabajando con lo que nos piden π₯ π¦ π₯2 + π¦2 π= + = π¦ π₯ π₯π¦ Elevando al cuadrado (π₯ + π¦)2 = (4)2 Aplicando binomio (π + π)2 = π2 + 2ππ + π 2 π₯ 2 + 2π₯π¦ + π¦ 2 = 16 π₯ β2 + π¦ 2 + 2π₯π¦ = 16 Reemplazando 8 + 2π₯π¦ = 16 π₯π¦ = 4 π₯2 + π¦2 8 π= = =2 π₯π¦ 4 15. Calcular: (π + π)(π3 β π 3 ) π= + π2 π2 + ππ + π 2 SoluciΓ³n Aplicamos diferencia de cubos π3 β π 3 = (π β π)(π2 + ππ + π 2 ) π=
(π + π) β (π β π)(π2 + ππ + π 2 ) + π2 (π2 + ππ + π 2 )
π = (π + π)(π β π) + π 2 Aplicando diferencia de cuadrados(π + π)(π β π) = π2 β π 2 π = π2 β π 2 + π 2 π = π2