# Engenharia

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• Publicado : 17 de janeiro de 2013

Amostra do texto
1)3+7x<8x-9
7x-8x<9-3
X>-6 S ] -6 ; +∞[

2)2-7<5x+3≤9
3-7<-5x ≤9 - 3
4<x ≤6 S : ] -∞, 4/5] U [6/5, +∞[
4/5<x ≤ 6/5

3) (x+6)(x-3) >0
X+6>0 X-3>0
X>-6 X>3 ] -∞, -6[ U ]3, +∞[

4) │5x-3│=7
5x-3=7
X=10/5
X=2

5) │7x-1│<│2x+5│
7x-1=2x+5 9x= -4
7x-2x=5+1 x= -4/9
5x=6
X=6/5

6) │9x+7│=-7
9x+7=-7
9x=-14NÃO É POSSIVEL RESOLVER POR O MODULO TER IGUALDADE POSITIVA
X=-14/9

7) 7X-2> -4
7x-2<4 X=-2/7
x< 6/7 S= (-∞, -2/7) U (6/7+∞)

8)
2x-5>3 2X-5<-3
2x>8 X<1
x>4

9) IABI =√[(Xb-Xa)²+(Yb-Ya)²]

IABI =√[(9-1)²+(9-3)²]
IABI =√[(8)²+(6)²]
IABI =√[64+36]
IABI =√[100]
IABI =10
10)
PA² = (x-4)²+(2+2)² = x²-8x + 32 [1]
PB² = (x-2)²+(2+8)² = x² -4x +104 [2]
Por ser equidistantes as quantidades devem ser iguais=
x² -8x + 32 = x² -4x +104 .......... (-x²)
-8x + 32 = -4x + 104 ......... (+4x)
-4x + 32 = 104 ........ (-32)
-4x = 72 .......... (/-4)
x = -18
11)
-1+3/2=2/2=1
1+5/2=6/2=3

Centro: (1;3)

12)
AB²=(3-1)²+ (y-0)²
AB²=9-3-3+1+y²
AB²=4+y²AC²=(3-5)²+(y-2)²
AC²=9-15-15+25+y²-2y-2y+4
AC²=8+y²-4y
AB²=AC²
4+y²=y²-4y+8
-4=-4y
Y=1
13)
{ 3x-2y=8 3X-2Y=8 3X-2Y=8 5x-6y=8
{ x+2y=8 X +2y=8 3.4-2y=8 5.4-6.2=8
{ 5x-6y=8 4x=16 -2y=-4 20-12=8
X=4 y=2 8=8

14-
R: X+2Y+3=0 Pr(0, -13/2) R(X+2Y+3=0)
S: X+2Y+13=0
X=0 S: 0+2Y+3=0 d(P,r) |axₒ+byₒ+c|
Y=-3/2 √(a²+b²)
dPr =|1.0+[(-3/2).2]+13 =
√[1²(+2)²]
dPr= 1-3+13
√5
dPr= 10 . √5 = 10√5 = 2√5
√5 √5 5

15-
r= 3x-4y=0 (x,2x)

d(P,r) |axₒ+byₒ+c|
√(a²+b²)
D= |3.x+(-4-2x) |
√9+16 -5x=15 -5x=-15
D= 3x-8xx=-3 x=3
√25 (-3,-6) (3,6)
3=| 5x |
5
15=| 5x|
16-
X+y+0-(1+6y)=0
X+y-6y-1=0 y=max+b
5y=x-1 5=-5.(5)+b
Y=x-1 b=30
5
M=1/5
Ma= -5

A(5,5)
B(1,0)
C(6,1)

17-
= x+2
-3+2=-1
-1+2=1 a imagemf(x)=x+2={-1,1,2,4}
0+2=2
2+2=4

18- 5x+3 D(F):{ X E R /X ≠-4,4}
X²-16

19-√(5-3x) D(F):{ X E R /X ≤ 1}

20- √(x-4) + 1__
√(x-2) D(F):{ X E R /X ͕>2}

21-
Log X+1 (X²+3X-18) D(F):{ X E R /X > 2}

22-
1_ . 6+ 1-4-1=
2
F(6)+g(-2)=3+1+4-1=
F(6)+g(-2)= 7

23- lim f(x)→ _1_ = -1
x→-1⁻¹ X

24- lim f(x)→ _1_ = -1x→-1⁺¹ X

25- lim f(x)→ x²=1
x→1⁻¹

26- lim f(x)→2-x= 1
x→1⁺

27- lim f(x)→ _1_ = -∞
x→0⁻ X

28- lim f(x)→ x²= 0
x→0⁺

29- lim f(x)→ x²= 0
x→0

30- lim _1_ = o⁺
x→+∞ 2 ͯ

31- - lim _1_ = o⁻
x→-∞ 2 ͯ

32- lim x-8__
x→8 ᶾ√(x)-2

lim y³-8___ = y³-8__
y→2ᶾ√(y³)-2 y-2

lim (y²+2y+4) = 4+4+4= 12
y→2

33- lim ᶾ√(x²)- 2ᶾ√(x)+1 fazendo x=y³ sendo x →1→3→1 então y³→ 1 logo ᶾ√1=1
x→1 (x-1)²
lim ᶾ√(y³) ²- 2ᶾ√(y³)+1
y→1 (y³-1)²
lim ᶾ√(y⁶) - 2ᶾ√(y³)+1
y→1 (y³-1)²
lim y² - 2y+1
y→1 (y³-1)²
lim (y-1)²
y→1 (y³-1)²
lim (y-1)(y-1)y→1 (y-1)(y²+y+1)
lim (y-1)__ = 0 = 0
y→1 (y²+y+1) 3

34- lim 3-√ (5+x) = 3-√ (5+x) . 1 +√ (5+x) . 3+√ (5+x) = 1+ √5-4 = 2 = 1/3
x→4 1 -√ (5+x) 1 -√ (5+x) 1 +√ (5+x) 3+√ (5+x) 3+√5+4 6

35- lim √(1+x) – 1 fazendo t= 1+x
x→0 ᶾ√(1+x) -1 se x→0 então t=1 e y⁶=t então y⁶=1 e...