Question

Me ajuda ai gente nessa duas ultima questão com o calculo por favor ?

Answer 1

1) $1+\left(\dfrac{1}{5}+\dfrac{1}{3}\right)\div\left(\dfrac{3}{5}-\dfrac{1}{15}\right)=1+\dfrac{3+5}{15}\div\dfrac{9-1}{15}=1+\dfrac{8}{15}\div\dfrac{8}{15}=1+1=2$

Letra C

2) $\sqrt{\dfrac{36}{25}\div\left(\dfrac{1}{2}\right)^4}=\sqrt{\dfrac{36}{25}\div\dfrac{1}{16}}=\sqrt{\dfrac{36}{25}\cdot\dfrac{16}{1}}=\sqrt{\dfrac{36\cdot16}{25}}=\dfrac{6\cdot4}{5}=\dfrac{24}{5}$

Letra D

Answer 2

a)
$1+\left(\frac{1}{5}+\frac{1}{3}\right) \div \left(\frac{3}{5}-\frac{1}{15}\right)=\\ \\ 1+\left(\frac{3+5}{15}\right) \div \left(\frac{9-1}{15}\right)=\\ \\ 1+\frac{8}{15}\div\frac{8}{15}=1+\frac{8}{15}*\frac{15}{8}=1+1=2$

b)

$\sqrt{\frac{36}{25}\div\left(\frac{1}{2}\right)^4} =\\ \\ \sqrt{\frac{36}{25}\div \frac{1}{16}} =\\ \\ \sqrt{\frac{36}{25}*\frac{16}{1}}=\sqrt{\frac{576}{25}}= \frac{24}{5}$