ecuaciones 1º
TRANSCRIPT
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ECUACIONES DE
PRIMER GRADO
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IGUALDAD ALGEBRAICA:Una igualdad algebraica posee números y letras (variable o incógnita)
ECUACIÓN:Es una ecuación porque es verdadera para un solo valor de “x”.
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ECUACIONES DE PRIMER GRADO:Una ecuación de primer grado con una incógnita es una igualdad algebraica que se puede reducir a la forma: ax + b = 0Donde:
a y b son número reales y a 0
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APRENDIZAJE PRELIMINAR
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Problema 01:Resuelve:
x + 1 = 3x – 7
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Problema 02:Resuelve:
2x + 3(x – 2) = 24
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Problema 03:Resuelve:
(x – 1)2 – (x – 2)2 = 10
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PROBLEMASPROPUESTOS
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Problema 01:Resuelve:3(5 – x) = 4(x – 5)6a – (10 – a) = 20 – (a – 2)20(x–2)–15(2x–3) = 20(5x–7)–75138 – 2(6x – 3) = 15(2x + 4)3(x – 7) + 9 = 4(5 – x) + 6x
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Problema 02:Resuelve:2(x + 1) + 3(x – 2) = 4(2 – x) + 68(x – 1) – 5(x + 3) = 2(x + 8)5x–[3x–2(x–5)] = [3(4–x)–5(x–2)+4][7(3–x) + 8(x–4)]–[2–3(x+2)+x] = 8
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Problema 03:Resuelve:(x + 3)2 = (x – 2)2 – 5(x + 2)2 = 32 + (x – 2)2
(x + 5)2 - (x – 5)2 = 5(2x + 4)(x – 3)2 = (x – 3)(x – 4)(x – 7)(x + 7) = (x – 5)(x + 10) – 4
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ACTIVIDADNº 01
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Problema 01:Resuelve:a)12x – 15 = 5x + 13b)17x + 9 = 6x +31c) 5x + 3(x – 1) = 13d)2(x + 1) + 3(x – 2) = 4(x + 1)e) 35(34 – x) = 70(7x – 13)
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Problema 02:Resuelve:a)(x – 3) – (x – 5) = 5(x – 1) – 3(x + 1)b)3x – (15 – x) = 17 – (x + 4) – xc)2(3 – x) – 3(x – 1) = 5(x+2) – (2x+9)d)[2–3(x+1)]+5(x–2) = {3[2–(x+3)]+1}e)8(x+1)–5(x+3) = 2x–{(x–3)–(5+x)}
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Problema 03:Resuelve:(x + 9)2 = (x – 9)2 + 72(x + 4)(x + 1) = (x + 2)2
4(x + 2) + 5 = 2(x + 7) + x(x + 5)2 = (x + 4)2 – 7(x + 3)2 = 64 + (x – 3)2