mate tarea 2º
TRANSCRIPT
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MATEMÁTICA
VALOR NUMÉRICO
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VALOR NUMÉRICO DE UN POLINOMIOEs el valor que toma el polinomio cuando a sus variables se les asignan valores particulares.
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APRENDIZAJEPREVIO
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EJEMPLO Nº 01
CLAVES
b) 33c) 32 d) 34e) 35
a) 30
Si:P(x) = 2x + 5
Halla P(14)
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EJEMPLO Nº 02
CLAVES
b) 101 c) 102d) 103e) 106
a) 100
Para:a = 5 b = 4
Halla el valor numérico de:
5ab + 3b – 2a
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EJEMPLO Nº 03
CLAVES
b) 43c) 42 d) 44e) 45
a) 40
Si:P(x) = 2x + 3
Calcula: P(P(P(3)))
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APLICO LO APRENDIDO
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PROBLEMA Nº 01Para:
a = 7 b = 5Halla el valor numérico de:a) 8ab + 6b – 3ab) 6a3 – 3b3 + ab
c) 12b4 + 5a2 – 4a + 5bd) 36b + 24a – 2a2 + 3b2
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PROBLEMA Nº 02Calcula el valor numérico de los
siguientes polinomios para:x = 4 , y = 5 , z = 3
a) P(x, y) = 7x2 – 10y3
b) P(x, y, z) = 8z4 + 3x3 – y2
c) P(x, y, z) = 3z2 + 3x3 + 3y5
d) P(x, y, z) = y – 3x + 7z
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PROBLEMA Nº 03Si:
P(x, y, z) = 18x2 – 23y3 + 7z4
Calcula:a) P(1, 3, 2) – P(0, 1, 2)b) P(3, 4, 5) – P(1, 2, 3)c) P(5, 6, 4) – P(1, 0, 3)d) P(7, 4, 3) – P(2, 0, 1)
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CAMBIO DE VARIABLE EN UN POLINOMIO Consiste en reemplazar una nueva variable por otra, de tal manera que se obtenga un nuevo polinomio en función de la nueva variable.
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APRENDIZAJEPREVIO
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EJEMPLO Nº 01
Sea:P(a) = 3a – 1
Halla: P(x + 2)
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APLICO LO APRENDIDO
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PROBLEMA Nº 01
Si:P(x) = 2x – 3
Halla: P(z)
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PROBLEMA Nº 02
Si:P(n) = 2n – 3
Halla: P(x + 2)
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PROBLEMA Nº 03
Si:P(y) = 3y + 2
Halla: P(5x) – 5P(x)
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PROBLEMA Nº 04
Si:P(3x - 2) = 6x + 1
Halla: P(m)
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PROBLEMA Nº 05
Si:P(4x – 1) = 8x – 7
Halla: P(a + 1)
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PROBLEMA Nº 06
Si:P(x) = 5x + 3Q(x) = x – 7
Halla: P(Q(a))
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PROBLEMA Nº 07
Si:P(y) = 2y + 6
Halla:P(4x) – 4P(x)
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ACTIVIDAD Nº 1
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PROBLEMA Nº 01
Si:P(x) = 5x5 – 3x2 + 7x + 15
Halla: P(–1)
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PROBLEMA Nº 02
Si:P(x) = 5x2 + 7x – 12
Halla:
22P
PPM
)2(
2)0(
)200(P)1(
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PROBLEMA Nº 03
Si:P(x) = x4 – 2x2 + 1
Halla:P[P[P[P[P(1)]]]]
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PROBLEMA Nº 04
Si:P(x) = 7x4 – x2 – 3x – 6
Halla: P(1)
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PROBLEMA Nº 05
Si:P(x) = 2x – 1
Halla:M = P[P[P(0)]]
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PROBLEMA Nº 06
Si:P(a) = 2a + 6
Halla:P(4x) – 4P(x)
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PROBLEMA Nº 07
Si:P(2x - 1) = 8x + 4
Halla: P(a)
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PROBLEMA Nº 08
Si:P(4x + 2) = 4x – 6
Halla: P(y – 1)
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PROBLEMA Nº 09
Si:P(x) = 2x + 1Q(x) = x – 3
Halla: Q(P(a))
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PROBLEMA Nº 10
Si:P(2x – 1) = 8x + 4
Halla: P(9)