3x......................6
-------- = 1 + ---------....sacamos comundenominador
x - 2..................x - 2
3x................(x - 2) + 6
-------- = ------------------
x - 2..................x - 2
3x................x + 4
-------- = ------------------
x - 2..................x - 2
3x (x - 2) = (x + 4) (x - 2).....(x -2) los cancelas
3x = x + 4
3x - x = 4
2x = 4
x = 4/2
x = 2
4............................10...................1
-------------+------------------= ----------------
2x - 3..............4x² - 9................2x + 3
(2x + 3)*4 + 10..................1
--------------------- = ----------------
4x² - 9 ........................2x + 3
8x + 12 + 10 ............1
--------------------- = -----------
(2x - 3)(2x + 3)..........2x - 3
(8x + 22)(2x - 3)
--------------------- = 1....cancelamos (2x - 3)
(2x- 3 ) (2x + 3)
8x + 22 = 1( 2x + 3)
8x + 22 = 2x + 3
8x - 2x = 3 - 22
6x = 19
x = 19/6
l3x + 2l = 5
entonces
3x + 2 = 5.......................-(3x + 2) = 5
3x = 5 - 2.........................3x = - 5 - 2
x = 3/ 3..............................x = - 7 /3
x = 1..................................x = -7/3
a) -5<= 4-3x/2<2...pasamos el 2 multiplicando
-10<= 4-3x < 4
-10 - 4 <= -3x < 4 - 4
- 14/ -3>= x > 0/-3....cambia el signo orde porque pasamos dividiendo un número negativo!
14/-3>= x > 0
el conjunto de solucion es ( 0; 14/3 ]
b) (2x - 3) (4x + 5) <= (8x + 1) (x – 7)
8x² + 10x - 12x - 15 <= 8x² - 56x + x - 7
8x² - 2x - 15 <= 8x² - 55x - 7
-2x + 55x <= - 7 + 15
53x <= 8
x <= 8/53
espero que te sirva, salu2!!!
