Question

Me ajudem apenas na questão 9 pfvr

Answer 1

Explicação passo-a-passo:

9)

a)

$12 {x}^{2} .3 {x}^{3} = 36 {x}^{2 + 3} = 36 {x}^{5}$

b)

$- a.( - 5 {a}^{5} ) = - 1 {a}^{1}.( - 5 {a}^{5} )= 5 {a}^{1 + 5} = 5 {a}^{6}$

c)

$\frac{1.5 {a}^{2}.b }{ - 5a. {b}^{3} } = \frac{ \frac{15}{10} {a}^{2} .b}{ \frac{ - 5a. {b}^{3} }{1} } = - ( \frac{15}{10} {a}^{2} .b \: \times \frac{1}{ 5a. {b}^{3} } ) = \\ \ \frac{ \frac{3}{10}a }{ {b}^{2} } = 0.3 a \: \div \: {b}^{2}$

d)

$( \frac{ {x}^{2} y}{3} )( \frac{2x {y}^{2} }{3} ) = ( \frac{( {x}^{2}y)(2x {y}^{2} }{3 \times 3} ) = \\ \frac{ 2{x}^{3} {y}^{3} }{9}$

e)

$\frac{32 {x}^{6} }{4 {x}^{4} } = 8 {x}^{6 - 4} = 8 {x}^{2}$

f)

$\frac{ (\frac{24 {x}^{3} {y}^{3} }{5}) }{ (\frac{ -8 {y}^{2} }{15} )} = \frac{24 {x}^{3} {y}^{3} }{5} \times \frac{ 15}{ - 8 {y}^{2} } = - 9 {x}^{3} y$