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Multiplicación algebraica

Producto de Binomios con término común

2
2
(x + a)(x + b) = x + xb + ax + ab  (x + a)(x + b) = x + (a + b)x + ab
Ejemplos:
2
2
• (x + 3)(x + 4) = x + 7x + 12 • (x – 8)(x + 6) = x – 2x – 48
2
2
• (x – 5)(x – 3) = x – 8x + 15 • (x – 2)(x + 9) = x + 7x – 18
Binomio y trinomio al cuBo

2
3
(a + b) = a + 3a b + 3ab + b 3 (a – b) = a – 3a b + 3ab – b 3
2
3
3
2
2
3
3
= a + b + 3ab(a + b) = a – b – 3ab(a – b)
3
3
3
(a + b + c) = a + b + c + 3(a + b)(b + c)(a + c)
3
3
3
3
3
3
3
= a + b + c + 3(a + b + c)(ab + bc + ac) – 3abc
2
2
2
3
3
3
= a + b + c + 3a (b + c) + 3b (a + c) + 3c (a + b) + 6abc
Problema 3 Resolución:
2
2
2
2

Si a + b + c = 10 • (a + b + c) = a + b + c + 2(ab + bc + ac)
2
2
2
2
a + b + c = 38 10 = 38 + 2(ab + bc + ac)  ab + bc + ac = 31
3
3
3
a + b + c = 160 • (a + b + c) = a + b + c + 3(a + b + c)(ab + bc + ac) – 3abc Recuerda
3
3

3

3

3

calcule abc. 10 = 160 + 3(10)(31) – 3abc  abc = 30
Rpta.: 30 Identidad de Lagrange Resuelve problemas de regularidad, equivalencia y cambio (Álgebra)
2
2
2
2
(a + b )(x + y )
2
igualdades condicionales = (ax + by) + (ay – bx) 2
a. Si a + b + c = 0 se cumple:
2 2
2 2
2
2
2
2 2
2
• a + b + c = –2(ab + bc + ac) • (ab + bc + ac) = a b + b c + a c
2
4
4
3
3
2 2
3
4
2
• a + b + c = 3abc • (a + b + c ) = 2(a + b + c )
2
7
7
5
5
2
b
b
2
b
b
3
b
b
5
5
3
2
• a + + c 2 ⋅ a + + c 3 = a + + c 5 • a + + c 2 ⋅ a + + c 5 = a + + c 7
2 3 5 2 5 7
b.
2
2
2
Si a + b + c = ab + bc + ac  a, b, c   a = b = c
2n
2n
n n
n n
n n
2n
En general: si a + b + c = a b + b c + a c  a, b, c , n N  a = b = c
c. Si a + b + c = 0  a, b, c   a = b = c = 0
2
2
2
Problema 4 Resolución:
Si a + b + c = ab + bc + ac, Si a + b + c = ab + bc + ac  a = b = c
2
2
2
2
2
2
siendo a, b y c reales, ab bc ca cc cc cc
+
2
2
+
+
+
2
2
2
2
+
ab bc ca abc = ccc = 3
2
+
2
2
calcule .
abc Rpta.: 3
Álgebra 5 - Secundaria 11