ejercicio 130 libro de baldor
5
8.- = = = 9. _2x+1_ – __x 2 ___+ _2x__ 12x+8 6x+x-2 16x-8 m.c.m= 8 (3x+2) (2x-1) 12x+8= 4(3x+2) 6x+x-2= (3x+2)(2x-1) 16x-8= 8(2x-1) = 2(2x+1)(2x-1) -8x 2 + 2x (3x+2) 8 (3x+2) (2x-1) = 8x 2 – 2 - 8x 2 + 6x 2 + 4x 8 (3x+2) (2x-1) = __6x 2 + 4x – 2__ 8 (3x+2) (2x-1) =__2(3x 2 +2x -1)_ =__3x 2 + 2x – 1__ 8 (3x+2) (2x-1) 4 (3x+2) (2x-1) 10. _1 - ____1 ____ + _1_ mcm=ax (a+x) ax a 2 + ax a+x a+x-x+ax = _ax+a__ = _a(x+1)_ = ax(a+x) ax(a+x) ax(a+x)
-
Upload
ivan-lobato-baltazar -
Category
Documents
-
view
11.640 -
download
6
Transcript of ejercicio 130 libro de baldor
8.-
=
=
=
9. _2x+1_ – __x2___+ _2x__
12x+8 6x+x-2 16x-8
m.c.m= 8 (3x+2) (2x-1)
12x+8= 4(3x+2)
6x+x-2= (3x+2)(2x-1)
16x-8= 8(2x-1)
= 2(2x+1)(2x-1) -8x2 + 2x (3x+2)
8 (3x+2) (2x-1)
= 8x2 – 2 - 8x2 + 6x2 + 4x
8 (3x+2) (2x-1)
= __6x2 + 4x – 2__
8 (3x+2) (2x-1)
=__2(3x2 +2x -1)_ =__3x2 + 2x – 1__
8 (3x+2) (2x-1) 4 (3x+2) (2x-1)
10. _1 - ____1____ + _1_ mcm=ax (a+x)
ax a2 + ax a+x
a+x-x+ax = _ax+a__ = _a(x+1)_ =
ax(a+x) ax(a+x) ax(a+x)
__x+1__
ax(a+x)
11.-
m. c. m.
12.-
=
13.-
+
-
=
+
-
= 1. (a+6)(a-4) M.C. D.
2. (a-4)(a+2) = (a+2)(a-4)(a+6)
3. (a+6)(a+2)
=
=
14.-
xy=(x)(y)
15.-
a3 + a+3 - a-1 m.c.m: (a+1) ( a2-a+1)
a3+1 a2-a+1 a+ a3+1= (a+1) ( a2-a+1)
a2-a+1= NO HAY= a2-a+1
a+1=NO HAY= a+1
a3 + a+3 - a-1 =
(a+1) ( a2-a+1)
a3 + a2+4a+3 -(a3 -2a2 -2a-1) =
(a+1) ( a2-a+1)
a3 + a2+4a+3 -a3 +2a2 -2a + 1 =
(a+1) ( a2-a+1)
3a2 + 2a +4 =
(a+1) ( a2-a+1)
R= 3a2 + 2a +4
a3+1
