a/b
28
1 − 13
−6 =
33
2
2
0,25 =
25 1
=
100 4
!
#
$
"
%
$
$
5
= 0,625
8
1
= 0,33...
3
e
13
= 1,181818...
11
2 ≅ 1,41421356
&
$
'
(
2,5
)
π ≅ 3,14159265
e
2
π
*
*
+
,
'
a > b. +
(
,
5 > 12
b
− 12 < 0
a
-
a < b.
(
.
-
e
− 8,2 < −2,4
a
b
6 7
<
7 8
.
&
≥
/ (
6 48
=
7 56
"
' $
− 3 ≥ −4
.
$
≤
/ (
− 3 ≥ −3
0
− 4 ≤ −3
" "
.
− 4 ≤ −4
e
>, <, ≥ e ≤
/ (
' $
1
2
'
1
"
a≤ x<b
&
"
$
"
'
'
"
/
3
"
' $
"
&
$'
1
(
"
"
(
≤
$ 1
[
a
5
$
(
$'
(
≤ (
5
≤
$ 1
)
b
"
<
(
(
a
]
b
"
[
a
4
)
b
a≤ x<b
$
"
1
b
"
5
&
/ (
a
[
a
&
1
4
4
"
&
(
5
"
(
a
< (
)
b
<
$ 1
]
b
≤ (
5
"
≥
$
5
"
[
a
5
"
≤
$
5
"
$
7
]
a
8 ' . * 95
)
a
"
2
'
"
"
' $
'
(
a-2
]
a
3
(
a
-2
]
b
3
4
:x≤3
( : x > −2
;
a ) x < −1
6
(
a
<>≥
+
$
' $
b) − 1 ≤ x ≤ 2
:
>
:−2< x ≤3
"
'
c) x > 2
<
=
$'
& "
(
"
'
"
(
?
"
"
'
−3 = 3
"
!
& "
3 =3
"
(
x.
(
0 =0
5−9 = −4 = 4
e
!
& "
(
'
'
'
'
A
"
?
(
$'
$
4
@
(
@
@
>
:
'
Solução:A distância de dois números é o valor absoluto da diferença entre eles. Assim,
?
B −2−3 = −5 = 5 .
@
.
'
4
"
(
(
' (
?
4
@
" "
"
"
(
$
+ '
x − 1 ≤ 3.
/
Solução: Em termos geométricos, números x tais que x − 1 ≤ 3 são aqueles cujo a distância de 1 é
igual ou menor que 3. Como mostra a figura abaixo, esses números satisfazem a
desigualdade − 2 ≤ x ≤ 4.
"
'
x −1 ≤ 3
− 3 +1 ≤ x −1+1 ≤ 3 +1
' (
$
,
− 3 ≤ x −1 ≤ 3
−2≤ x≤4
4
'
(
(
(
"
(
.
'
#
'
$
$
%
a = a⋅a
n
"
%
8
%
C '
%
a = a⋅a
n
a
a −x =
"
a0 = 1
1
ax
A
0
'
a )91 / 2
b)27 3 / 2
c)8 −1 / 3
−3 / 2
e)5 0
+
4
a )91 / 2 = 9 = 3
b)27 2 / 3 =
( 27 ) = 3
2
3
2
=9
= 3 (27 ) = 3 729 = 9
2
c)8 −1 / 3 =
d)
1
100
1
1/ 3
8
=
1
3
8
=
1
2
−3 / 2
= 100 3 / 2 =
( 100 ) = 10
3
3
= 1000
e)5 0 = 1
"
&
"
D
D
D
(
'
a r a s = a r+s
s
%
$
"
C
1
d)
100
a
a r / a = a r − s (a ≠ 0)
(a r ) s = a rs
an/m =
( a)
m
n
= m an
&
'
A
0
( ) =2
3
a) 2 −2
b)
33
31 / 3 3 2 / 3
c) 2
A
(
)
(8 −1 / 4 ) = 2 7 / 4 2 −3 / 4 = 2 7 / 4−3 / 4 = 2 ' = 2
"
40
'
a5
= an
a2
a5
= a 5− 2 = a 3 , n = 3
a2
( ) =a
(a ) = a ,5n = 20
( : an
A
:
=
?
:
>
1
1
=
6
64
2
3
3
33
= 1 / 3+ 2 / 3 = 1 = 3 2 = 9
3
3
−6
( )
7/4
'
5
20
n 5
%
ou n = 4
5n
$
8
#
(
"%
. $
4
(
$
1
$
"
4
0
"
4
Lei da Distributividade: Para quaisquer números reais a,b e c,
ab + ac = a (b + c)
&
'
%
'
$
$
4
40
3x 4 − 6 x 3
: 8
+
4
A
(
" /"
"
E
+
+
4
" %
F
3 x 4 − 6 x 3 = 3 x 3 ( x − 2)
"
10(1 − x) 4 ( x + 1) 4 + 8( x + 1) 5 (1 − x) 3
$
&
2(1 − x) 3 ( x + 1) 4 A
" /"
$
" %
10(1 − x) 4 ( x + 1) 4 + 8( x + 1) 5 (1 − x) 3 = 2(1 − x) 3 ( x + 1) 4 [5(1 − x) + 4( x + 1)]
A
/"
$
1
4
$
(
1 '
$
10(1 − x) 4 ( x + 1) 4 + 8( x + 1) 5 (1 − x) 3 = 2(1 − x) 3 ( x + 1) 4 (5 − 5 x + 4 x + 4) = 2(1 − x) 3 ( x + 1) 4 (9 − x)
3
0
$
$
2
2
( x + 1)
( x + 1) + x 2 / 3 ( x1 / 3 )
2/3
3
3
=
+x =
= >
x1 / 3
x1 / 3
$
#
$
&
%
+ '
'
+
'
/"
$
'
(
$
G
$
4
$
'
4( x + 3) 4 ( x − 2) 2 − 6( x + 3) 3 ( x − 2) 3
( x + 3)( x − 2) 3
$
$
-
:
$
0
4
$
4
2 −1 / 3
x ( x + 1) + x 2 / 3
3
+
" "
2 −1 / 3
x ( x + 1) + x 2 / 3
3
:8
&
0
(
4( x + 3) 4 ( x − 2) 2 − 6( x + 3) 3 ( x − 2) 3 2( x + 3) 3 ( x − 2) 2 [2( x + 3) − 3( x − 2)]
=
=
( x + 3)( x − 2) 3
( x + 3)( x − 2) 3
2( x + 3) 3 ( x − 2) 2 (2 x + 6 − 3 x + 6) 2( x + 3) ( x − 2) 2 (12 − x)
=
=
( x + 3)( x − 2) 3
( x + 3)( x − 2) 3
3
3
'
( x + 3)( x − 2) 2
$
4( x + 3) 4 ( x − 2) 2 − 6( x + 3) 3 ( x − 2) 3 2( x + 3) 2 (12 − x)
=
x−2
( x + 3)( x − 2) 3
&
(
+
a1 + a 2 +
$
(
$
+ an
/
H
'
"
$
"
j =1
> '
!
:
1
a1 + a 2 +
: 2
I
A soma dos números a1 + a 2 +
Somatório
4
4
+ an =
2
' '
!
+ a n pode ser representada por
n
j =1
aj
!
'
: 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64
( : (1 − x1 ) 2 ∆x + (1 − x 2 ) 2 ∆x +
+
+ (1 − x15 ) 2 ∆x
4
: 3
J
1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 =
$
8
4
j2
j =1
(: &
(1 − x j ) 2 ∆x .
j =1
j =8
.
: A
4
:
!
( j 2 + 1)
'
j =1
+
3
b)
(−2) j
j=1
4
4
( j 2 + 1) = (12 + 1) + (2 2 + 1) + (3 2 + 1) + (4 2 + 1) = 2 + 5 + 10 + 17 = 34
j =1
3
(−2) j = (−2)1 + (−2) 2 + (−2) 3 = −2 + 4 − 8 = −6
j =1
%
$
)
2
&
$
I
'
&
*
a 0 , a1 , a 2 ,
, an
'
)
(
I
1
'
'
C
"
%
)
K
*
$
"
$
>
'
$
&
'
x 2 − 2x − 3
"
$
: &
I
I
(
G
"
$
I
4
$
(
I
+
I
#
I
I
3 x 5 − 7 x 2 + 12
I
&
4
*
40
$
: 8
+ an x n
#
'
I
$
$,
1
an ≠ 0
"
+
a 0 + a1 x + a 2 x 2 +
$
"
I
&
(
'
"
(
"
$
$
I
$
x 2 − 2 x − 3 = ( x + a )( x + b)
?
( x + a )( x + b) = x 2 + (a + b) x + ab ?
(
(
"
x 2 − 2 x − 3 = x 2 + (a + b) x + ab ou seja, tais que a + b = −2 e ab = −3
"
1,−3;−3,1;−1,3;3,−1
?
9
9
3
a = 1 e b = −3
.
4
'
$
"
$
$
(
: 8
4
a = −3, b = 1.
x 2 − 2 x − 3 = ( x − 3)( x + 1)
(
+
9
2 x 2 + x − 10
I
$
3
"
2 x 2 + x − 10 = (ax + b)(cx + d ) (lei da distributividade)
= (ac) x 2 + (bc + ad ) x + (bd )
ac = 2
bc + ad = 1
bd = −10
A
"
"
1
"
(
*
1
J
1
3
2,1;1,2;−2,−1;−1,−2
" 2,−5;5,−2;1,−10;10,−1;−2,5;−5,2;−1,10;−10,1
* J B
/"
$
# "
(
$
a = −2, b = −5, c = 1 e d = 2 de modo que 2 x 2 + x − 10 = (2 x + 5)( x − 2)
>& (
"
4
#
"
(
I
: 8
4
(
:
(
'
L:
x3 − 8
I
$
23 = 8
(
-
$
x−2
"
$
?
x 3 − 8 = ( x − 2)( x 2 + ax + b)
( x − 2)( x 2 + ax + b) = x 3 + (a − 2) x 2 + (b − 2a ) x − 2b
A
&
a = −2, b = −5, c = −1 e d = 2 &
$
"
$
+
(
$
a − 2 = 0 b − 2a = 0
"
$ -
%
40
"
e
$
2b = 8
a=2 e
b=4
x 3 − 8 = ( x − 2)( x 2 + 2 x + 4)
.
3
*
x 2 + 2x + 4
#
$
$
I
I
'
$
-
.
'
$
1 '
>
4
"
$
4
:
$
$
1
-
:
Diferença de dois quadrados: para dois números reais a e b quaisquer,
a 2 + b 2 = (a + b)(a − b)
: 8
+
x5 − 4x3
I
4
$
'
3
'
" %
x2 − 4 >
$
x 5 − 4 x 3 = x 3 ( x 2 − 4)
(
$
4
:
x 5 − 4 x 3 = x 3 ( x + 2)( x − 2)
$
.
+
$,
%
$,
4
x=2
"
4
4
"
"
x 3 − 6 x 2 + 12 x − 8 = 0
4
"
(
2 3 − 6(2 2 ) + 12(2) − 8 = 8 − 24 + 24 − 8 = 0
C
'
40
.
(
$
: ;
"
"
$
4
$
a = 0 e b = 0, ou a = b = 0
"
4
x 2 − 3 x = 10
"
"
Solução:Em primeiro lugar, subtraímos 10 de ambos os membros para obter
x 2 − 3 x − 10 = 0
3
'
$
I
(
( x − 5)( x + 2) = 0
( x − 5)( x + 2) !
A
$
40
x = 5 (anulando o primeiro fator) e x = −2 (anulando o segundo fator)
4
: ;
+
$
4
"
;
-
1 2
−
=0
x x2
1−
4
$ 40
x2 − x − 2
=0
x2
x2
x
2
− 2 − 2 = 0 ou
2
x
x
x
8
( x + 1)( x − 2)
=0
x2
I
A
!
$
x = −1 e x = 2
40
$
2
2
+
4
4
$,
'
&
.
ax 2 + bx + c = 0 (para a ≠ 0)
$
'
40
%&
1
40
$
4
$
1
-
" ' ()*
#
A
$
/"
4
'
"
'
$
!("
4
40
!
Fórmula de Báskara para resolver equações do segundo grau: As
soluções da equação do segundo grau
ax 2 + bx + c = 0 (para a ≠ 0 )
são dadas pela expressão x =
&
b 2 − 4ac
$!
M
N
− b ± b 2 − 4ac
2a
1
4
'
'
#
"
(
/
(
#
$!
: ;
M
−3+ 5
= −0,38
2
"
4
'
e
x=
"
-
'
a = 1, b = 18 e c = 81 2
'
$!
− 18 ± 0
= −9
2
x2 + x +1 = 0
4
4
4
x=
$!
−3− 5
= −2,62
2
x=
"
4
'
a = 1, b = 3 e c = 1 2
'
(
N
b
#
2a
x 2 + 18 x + 81 = 0
4
4
: ;
M
x=−
-
/
x 2 + 3x + 1 = 0 =
4
x=
N
+
N
4
N
(
4
1
4
+
$!
"
"
: ;
M
M
$!
4
40
'
±
40
/
4
±
40
O
±
+
4
'
a = 1, b = 1 e c = 1 2
'
−1± − 3
A
2
$!
-
9
40
&
+
$,
2
40
40
. '
"
"
(
!'
/
2
/
1
P
" "
4
40
+
$ -
2x + 3 y = 5
x + 2y = 4
&
"
>
:
"
!'
"
/
$
:;
9
.
$
− x + 0 = 5 ou
(
x = −5
$
x = −5
2 y = 22
40
ou
x = −5
y = 11
3(−5) + 2(11) = −15 + 22 = 7
$
&
40
$
$
2 y 2 − x 2 = 14
"
x − y =1
'
"
'
3
1
y = 11
4(−5) + 3(11) = −20 + 33 = 13
4
x = −5
y = 11
$ -
'
(
− 9 x − 6 y = −21
3(−5) + 2 y = 7
4
4
8 x + 6 y = 26
+ $ -
4
(
$ -
(
"
+ (
$
(
40
:;
"
'
0
4
+
4
"
"
3x + 2 y = 7
'
"
"
"
+
.
(
"
4 x + 3 y = 13
4
'
"
"
/
"
(
4
"
'
"
!'
(
?
40
4
C
-
"
(
+
40
x = y +1
4
+
4
'
C
2 y 2 − ( y + 1) 2 = 14
2 y 2 − ( y 2 + 2 y + 1) = 14
2 y 2 − y 2 − 2 y − 1 = 14
y 2 − 2 y − 15 = 0
&
( y + 3)( y − 5) = 0
y = −3
y = −3
'
x − (−3) = 1 ou
4
x −5 =1
4
40
4
8 -
$
x = 6, y = 5
40
8 B
(
B H Q B )
Q B
x = −2
y=5
'
x = −2
ou
?
"
y=5
x = −2, y = −3
e
/
"
+
>) : 9 H B ) 9 H B *
>
: 9>
: B
J 9* B
*
$
40