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tarea aleatoria.pdf
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Author:
Alfredo Vásquez Mendoza
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x2 − 4x − 21 = 0
(x − 7)(x + 3) = 0 x−7= 0 x + 3 = 0 x = 7 = − −33 x = 9x2 +54x+79 1
(9x2 + 54x + 79) = 0( 19 ) =0 x + 6x + 79 9 6 2 b 2 (2) = ( 2) = 9 =0 x2 + 6x + 9 − 9 + 79 9 2 2 (x + 6x + 9) − 9 = 0 (x + 3)2 = 92 9
2
x + 3 = x =
√
2 3
√
2 3
x + 3 = −
−3
x + 3 = −
√
2
3 √
2 3
−3 3x + 2 = 6 x + 4
3x + 2 − 2 = 6x + 4 − 2 3x − 6x = 6x − 6x + 2 −3x = 2 − 13 (−3x) = 2(− 2(− 13 ) 2 = − − 3 x = −3x + 1 < 2 x + 5 −3x + 1 − 1 − 2x < 2 x − 2x + 5 − 1 −5x < 4 − 15 (−5x) < 4(− 4(− 15 ) x > − 45 x2 − 5x + 6 < 0 A = (xR|x2 − 5x + 6 < 0) xR 2 x − 5x + 6 < 0
xA ⇔ x 2 − 5x + 6 < 0
( −25 )2 = 25 4 ⇔ x 2 − 5x + 25 − 4 25 2 ⇔ x − 5x + 4 − ⇔ ( x − 25 )2 < 41 − 12 +
< x − 25 + 4 < x < 26 2 2 < x < 3
25 4 1 4
+ 6 < 0 < 0
5
5
4
2
< 25 +
1 2
|x2 − 4| = − = −22x + 4
A = (xR||x2 − 4| = −2x + 4) xA ⇔ |x2 − 4| = −2x + 4
xR
|x2 − 4| = −2x + 4 ⇔
−2x + 4 ≥ 0
y − x2 + 4 = −2x + 4
x2 − 4 = −2x + 4
x ≤ 2 y x(−x + 2) = 0 o´ (x + 4)(x − 2) = 0 x ≤ 2 { y x = 0 o´ x = 2 o´ x = −4 x = 0 x = 2 x = −4 |x + 3| ≤ 5 A = (xR|||x + 3| ≤ 5)
{
xA ⇔ |x + 3| ≤ 5
xR
|x + 3| = 5 5 ≥ 0 |x + 3| = 5 ⇔ {
y
−x − 3 = 5 o´ x + 3 = 5 5 ≥ 0 ⇔
y x = −8
o
x = 2
|x + 3| < 5 x + 3 < −5 x + 3 < 5 x > −8 x < 2 −8 ≤ x ≤ 2 [−8, 2] ab = ac
= 0, a
b = c = b ∗ 1 = b · (a · a−1 ) = ab(a−1 ) = ac(a−1 ) = c (a · a−1 ) = c (1) = c
(2x − y ) + ( x + y ) = 3x 2x − y + x + y = 3x 3x = 3x
b = c.
ab > 0 ab · 1 > 0(1) 1 · ab > 0 ab > 0(1) ab > 0
−ab > 0 (−1)ab > 0 ab < 0(−1) ab > 0 a−1 a ∗ a−1 = 1 a · a−1 = 1 a a
=1 1 =1 x|2, 4|, 2 ≤ x ≤ 4 x ≥ 2 x ≤ 4 x > 2 x = 0 x < 4 x = 4 2x > 4 2x = 4 2x + 3 > 7 2x + 3 = 7 2x + 3 ≤ 7 2(2) + 3 = 7 2(4) + 4 = 11
2x + 3 [7, 11]
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