gabaritomat

Intensivo revisão

01. d

02. e

03. c

x3 + 1
(x + 1 ) ( x 2 − x + 1 ) x = x +1
=
=
1
x2 − x + 1 x2 − x + 1 x + −1 x x
5
5
8
Logo, para x = 1,666... = , temos x + 1 =
+1= .
3
3
3
8
6+
6,888...
9 = 62 = 31 = p
=
2,444...
4
22 11 q
2+
9
Logo p + q = 31 + 11 = 42. x2 +

12
5
−
7 +3 8−3 7
12
( 7 − 3)
5
=
$
−
( 7 + 3) ( 7 − 3) (8 − 3 7 )

Temos

=
04. a

1 x $

(8 + 3 7 )
(8 + 3 7 )

12 7 − 36 40 + 15 7
= 18 − 6 7 − 40 − 15 7 = −22 − 21 7.
−
7−9
64 − 63

Sendo a = 100, b = 1 000 e x = 0,09, então ax ⋅ b x
= (100)0,09 ⋅ (1 000) 0,09 = (102) 9/100 $ (103)0,3 = 109/50 $ 109/10
,
= 109/50 + 9/10 = 1054/50 = 10108 .

05. e

I. Falsa. Para x = y = 1, ex/y = e1 = e e ex – ey = e1 – e1 = 0.
0,001
0,04
0,01
1$ 10 −3
4 $ 10 −2 1$ 10 −2
II. Falsa.
=
+
−
+
−
0,04
0,001
0,2
4 $ 10 −2
1$ 10 −3
2 $ 10 −1
1
1
=
⋅ 10–1 + 4 ⋅ 10 –
⋅ 10–1 = 0,25 ⋅ 10–1 + 40 – 0,5 ⋅ 10–1
4
2
= 0,025 + 40 – 0,05 = 39,975.
III. Falsa. Para x = y = 1, x + y = 1 + 1 = 2 e x + y
= 1+ 1 = 2 .

06. c

0,064 =

07. c

1# a # 2
1
1# a # 2
+ 1 1 1 & 1$
5
#
#
3#b #5
5 b 3

08. a

64
8
2 3
=
=d n
1 000 125
5
#a $

1 b #2$

1
1
+
5
3

#

a b #

x +y − z =0
(x + y − z) + (xy − 10) + (z − 7) = 0 + xy − 10 = 0 z −7=0
Portanto, x + y – z = 0.
2

4

6

1

MATEMÁTICA

2
3

09. c

I. Verdadeira. Temos:
1 3
1
1
3
3
2 1 dk + n = 3 + k + 3k $ + 3k $ 2 + 3 = 27 k k k k
1
1
1
+ k3 +
= 27 − 3 $ 3 = 18
+ 3 dk + n = 27 + k3 +
3
k k k3
Logo

k3 +

1

_ 3+

5 +

3− 5i

k3
II. Verdadeira.

= 18 = 3 2
2

= ( 3 + 5 )2 + 2 $ 3 + 5

$

3 − 5 + ( 3 − 5 )2

= 3 + 5 + 2 $ (3 + 5 ) $ ( 3 − 5 ) + 3 − 5 = 3 + 2 ⋅

4 + 3 = 10

III. Falsa. x 2 − 4x + 4
(x − 2)2
= |x − 2| +
= |x − 2| x −2 x −2
|x − 2|
1
x − 2=1
+
+x =3
= |x − 2| +
= 1+ x −2 x −2 x − 2! 0
1
= 0,... xyz... xyz... onde x, y e z são os 3 últimos