85 - yoquieroaprobar.es

4
Soluciones a las actividades de cada epígrafe
PÁGINA 85
Pág. 1
1 Simplifica las fracciones siguientes. Para ello, saca factor común cuando convenga:
a)
15x 2
5x 2(x – 3)
d) 9(x + 1) – 3(x + 1)
2(x + 1)
a)
b) 3(x – 1)
9(x – 1)
2
2
3
c) 3x 3– 9x 4
15x – 3x
2
2
e) 5x (x – 3) (x + 3)
15x (x – 3)
3
2
f ) x (3x – x 3)
(3x – 1)x
3
x–3
b) x – 1
3
2
+1
c) 3x 3(1 – 3x) = 1 – 3x = –3x
2
3x (5 – x) x (5 – x) – x + 5x
d) (x + 1)(9 – 3) = 6(x + 1) = 3
2(x + 1)
2(x + 1)
2
3
e) x(x – 3)(x + 3) = x (x – 9) = x – 9x
3
3
3
2
3
f) x · x (3x – 31) = x (3x – 1)3 = 1
(3x – 1)x
(3x – 1)x
2 Opera y simplifica.
2
b) 3 – 2x 2 + 8x – 4x
x+1
x +x
3
2
3
2
d) 5x + 15x – 10x + 215x + 2x
x+3
5x
a) 2 + 3 + x – 2
x 2x
x
c)
x2
2 – 7x + 3
–9 x–3
a) 4 + 3 + 2(x – 2) = 7 + 2x – 4 = 2x + 3
2x 2x
2x
2x
2x
2
2
b) 3 – 2x + 8x – 4x = 3x – 2x + 8x – 4x · x(x + 1) =
x + 1 x(x + 1)
x (x + 1) x(x + 1)
x(x + 1)
2
3
2
3
2
= 3x – 2x – 8x – 4x – 4x = –4x – 6x – 5x =
x(x + 1)
x(x + 1)
2
2
= –x (4x + 6x + 5) = –4x – 6x – 5
x(x + 1)
x+1
c)
2
2
– 7x + 3 =
– 7x(x + 3) + 3(x + 3)(x – 3) =
(x + 3)(x – 3) x – 3
(x + 3)(x – 3) (x + 3)(x – 3) (x + 3)(x – 3)
2
2
2
= 2 – 7x – 21x + 3x – 27 = –4x – 21x – 25
(x + 3)(x – 3)
(x + 3)(x – 3)
2
2
d) 5x (x + 3) – 5x (2x2+ 3) + 2x = 5x 2 – (2x + 3) + 2x = 5x 2 – 2x – 3 + 2x = 5x 2 – 3
x+3
5x
Unidad 4. El lenguaje algebraico
4
Soluciones a las actividades de cada epígrafe
3 Efectúa las siguientes operaciones y simplifica. Ten en cuenta las identidades notables:
2
a) x – 1 : (x – 1)
x
2
b) x (x – 2) : x – 4
x
x+2
2
c) x – 2x + 1 : x – 1
x
x
d) 6x 2 · x –33
x
e) 3x –2 3 · x (x2 + 1)
x
x –1
2
f ) 2x : 4x
x – 1 2x – 2
5
g) x + 5 ·
10 (x + 5)2
2
h) 2x · 6x3
3x 4x
2
i) 4x – 3 · 4x
8x – 6
2x
3x
j) 3x –2 3 ·
18(x
– 1)
x
a) (x + 1)(x – 1) : (x – 1) = (x + 1)(x – 1) · 1 = x + 1
x
x
x–1
x
x+2
=1
b) x(x – 2) : (x + 2)(x – 2) = x(x – 2) ·
x
(x + 2)
x
(x + 2)(x – 2)
2
2
c) (x – 1) : x – 1 = (x – 1) · x = x – 1
x
x
x
x–1
2
d) 6x (x3– 3) = 6(x – 3) = 6x – 18
x
x
x
e) 3(x –2 1) · x(x + 1) = 3
(x + 1)(x – 1) x
x
2
f ) 2x : (2x) = 2x · 2(x – 21) = 2 = 1
x – 1 2(x – 1) x – 1 (2x)
2x x
1
g) 5(x + 5) 2 =
2(x + 5)
10(x + 5)
3
h) 12x 4 = 1
x
12x
2
= 2x = x
i) 4x – 3 · (2x)
2(4x – 3)
2x
2
3x
= 9x 2 = 1
j) 3(x –2 1) ·
18(x – 1) 18x
2x
x
Unidad 4. El lenguaje algebraico
Pág. 2
4
Soluciones a las actividades de cada epígrafe
4 Opera y simplifica.
Pág. 3
a)
6x 2 :
5x + 5x
2
2x – 3 2x + 3
4x – 9
b)
x2
1–
x3 + x2
–
5x 2 – 25 5 (x + 1)(5x 2 – 25)
a)
6x 2
: 5x(2x + 3) + 5x(2x – 3) =
(2x + 3)(2x – 3)
(2x + 3)(2x – 3)
(
=
b)
)
2
2
6x 2
· (2x + 3)(2x – 3) = 6x
= 6x 2 = 3
(2x + 3)(2x – 3) 5x(2x + 3 + 2x – 3) 5x · 4x 20x
10
5(x + 1)x 2
(x + 1)(5x 2 – 25) –
5(x 3 + x 2)
–
=
5 (x + 1)(5x 2 – 25) 5(x + 1)(5x 2 – 25) 5(x + 1)(5x 2 – 25)
2
2
2
2
2
2
= 5x (x + 1) – (x + 1)(5x 2– 25) – 5x (x + 1) = 5x – 5x 2+ 25 – 5x =
5(x + 1)(5x – 25)
5(5x – 25)
2
= –5x 2 + 25 = – 1
5
5(5x – 25)
Unidad 4. El lenguaje algebraico