Question

1- Se a.b # 0, simplificar as expressões:

a) ( a⁵ . b³ )² =
( a⁻⁴ . b)⁻³

b) ( a³ . b⁻⁴ )³ =
( a⁻² . b²)

c) [( a² . b⁻³)²]⁻³ =

d) ( a⁻¹ + b⁻¹) . ( a+b)⁻¹ =

Answer 1

a)

$\left(a^{5}\cdot b^{3}\right)^{2}=\left(a^{-4}\cdot b\right)^{-3}\\ \\ \left(a^{5\cdot 2}\cdot b^{3\cdot 2}\right)=\left(a^{-4\cdot \left(-3\right)}\cdot b^{-3}\right)\\ \\ a^{10}\cdot b^{6}=a^{12}\cdot b^{-3}\\ \\ \\ \dfrac{a^{12}}{a^{10}}=\dfrac{b^{6}}{b^{-3}}\\ \\ \\ a^{12}:a^{10}=b^{6}:b^{-3}\\ \\ a^{12-10}=b^{6-\left(-3\right)}\\ \\ a^{2}=b^{6+3}\\ \\ a^{2}=b^{9}$


b)
$\left(a^{3}\cdot b^{-4}\right)^{3}=\left(a^{-2}\cdot b^{2}\right)\\ \\ \left(a^{3\cdot 3}\cdot b^{-4\cdot 3}\right)=\left(a^{-2}\cdot b^{2}\right)\\ \\ a^{9}\cdot b^{-12}=a^{-2}\cdot b^{2}\\ \\ \\ \dfrac{a^{9}}{a^{-2}}=\dfrac{b^{2}}{b^{-12}} \\ \\ \\ a^{9}:a^{-2}=b^{2}:b^{-12} \\ \\ a^{9-\left(-2\right)}=b^{2-\left(-12\right)} \\ \\ a^{9+2}=b^{2+12} \\ \\ a^{11}=b^{14}$


c)

$\left[\left(a^{2}\cdot b^{-3}\right)^{2}\right]^{-3}\\ \\ \\ \left(a^{2}\cdot b^{-3}\right)^{2\cdot \left(-3\right)}\\ \\ \left(a^{2}\cdot b^{-3}\right)^{-6}\\ \\ a^{2\cdot \left(-6\right)}\cdot b^{-3\cdot \left(-6\right)}\\ \\ a^{-12}\cdot b^{18}$


d)

$\left(a^{-1}+b^{-1}\right)\cdot \left(a+b\right)^{-1}\\ \\ \left(\dfrac{1}{a}+\dfrac{1}{b}\right)\cdot \left(\dfrac{1}{a+b}\right)\\ \\ \\ \left(\dfrac{1\cdot b\ +\ 1\cdot a}{ab}\right)\cdot \left(\dfrac{1}{a+b}\right)\\ \\ \\ \dfrac{b+a}{ab}\cdot\ \dfrac{1}{a+b}\\ \\ \\ \dfrac{\left(a+b\right)}{ab\cdot\left(a+b\right)}=\dfrac{1}{ab}$

Answer 2

a) $\frac{(a^5.b^3)^2}{(a^-^4.b)^-^3} = \frac{a^1^0.b^6}{a^1^2.b^-^3} =(a^1^0.b^6):(a^1^2.b^-^3)= a^-^2.b^9$

b) $\frac{(a^3.b^-^4)^3}{(a^-^2.b^2)}= [tex] \frac{ a^{9}. b^{-12} }{ a^{-2} .b^2} =(a^9. b^{-12}):( a^{-2} .b^2)= a^{11} . b^{-14}$

c) $[(a^2.b^-^3)^2]^-^3= [a^4.b^-^6]^-^3= a^-^1^2.b^1^8$

d) igual à respondida pelo Alexandro