Question

Ajudem aí por favor.........

Answer 1

Resposta:

Explicação passo a passo:

1)

a)

3⁸.3⁴/3⁵.3⁷=3⁽⁸⁺⁴⁾/3⁽⁵⁺⁷⁾=3¹²/3¹²=3⁽¹²⁻¹²⁾=3⁰=1

b)

2⁶:2⁴/2³=2⁽⁶⁻⁴⁾/2³=2²/2³=2⁽²⁻³⁾=2⁻¹=1/2

2)

2ˣ=1/32

Fatore 32

32 | 2

16 | 2

8 | 2

4 | 2

2 | 2

1 | 1

32= 2 . 2 . 2 . 2 . 2 = 2⁵

2ˣ=1/2⁵, bases iguais compare os expoentes

2ˣ=2⁻⁵

x= -5

Alternativa d)

b)

25ˣ=1/125

Fatore 125

125 | 5

25 | 5

5 | 5

1 | 1

125 = 5 . 5 . 5 = 5³

Fatore 25

25 | 5

5 | 5

1 | 1

25= 5 . 5  = 5²

25ˣ=1/5³

(5²)ˣ=1/5³

5⁽².ˣ⁾=5⁻³

2x= -3

x= -3/2

Alternativa d)

c)

(3ˣ)ˣ⁻¹=9

3ˣ⁽ˣ⁻¹⁾=3²

x(x-1)=2, aplique a distributiva:

x²-x=2

x²-x-2=0

$\displaystyle Aplicando~a~f\acute{o}rmula~de~Bhaskara~para~x^{2}-x-2=0~~e~comparando~com~(a)x^{2}+(b)x+(c)=0,~determinamos~os~coeficientes:~\\a=1{;}~b=-1~e~c=-2\\\\C\acute{a}lculo~do~discriminante~(\Delta):&\\&~\Delta=(b)^{2}-4(a)(c)=(-1)^{2}-4(1)(-2)=1-(-8)=9\\\\C\acute{a}lculo~das~raizes:&\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-1)-\sqrt{9}}{2(1)}=\frac{1-3}{2}=\frac{-2}{2}=-1\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-1)+\sqrt{9}}{2(1)}=\frac{1+3}{2}=\frac{4}{2}=2\\\\S=\{-1,~2\}$

Alternativa d)

d)

(25)²ˣ=√5

(5²)²ˣ=√5

5⁴ˣ=5¹/²

4x=1/2

x=1/8

Alternativa d)