Question

a) 3 /√7
b) 3 /3√7
c)
d) 4 /3√3
e) 2 /√5
f) 7 /5√2

Answer 1

Explicação passo-a-passo:

racionalização

a)

${3\over\sqrt{7} }={3\sqrt{7} \over\sqrt{7} .\sqrt{7} }={3\sqrt{7} \over\sqrt{49} }={3\sqrt{7}\over7}\\ \\  b)\\ {3\over3\sqrt{7} }={1\over\sqrt{7} }={\sqrt{7} \over\sqrt{7} .\sqrt{7}}={\sqrt{7}\over\sqrt{49}  }={\sqrt{7} \over7$

$d)\\ {4\over3\sqrt{3} }={4\sqrt{3} \over3.\sqrt{3} .\sqrt{3} }={4\sqrt{3} \over3\sqrt{9} }={4\sqrt{3} \over3.(3)}={4\sqrt{3} \over9}\\ \\ e)\\ {2\over\sqrt{5} }={2\sqrt{5} \over\sqrt{5} .\sqrt{5} }={2\sqrt{5} \over\sqrt{25} }={2\sqrt{5} \over5}\\ \\ f)\\ {7\over5\sqrt{2} }={7\sqrt{2} \over5.\sqrt{2} .\sqrt{2} }={7\sqrt{2} \over5\sqrt{4} }={7\sqrt{2}\over5.(2)}={7\sqrt{2}  \over10}$

Answer 2

Olá, boa noite☺

Resoluções.(Racionalização de denominador)

a)

$\dfrac{3}{\sqrt{7} }=\dfrac{3}{\sqrt{7} }\times\dfrac{\sqrt{7}}{\sqrt{7}} =\boxed{\dfrac{3\sqrt{7} }{7}}$

b)

$\dfrac{3}{3\sqrt{7} }=\dfrac{3}{3\sqrt{7} } \times\dfrac{\sqrt{7} }{\sqrt{7} }=\dfrac{3\sqrt{7} }{3 \times 7} =\boxed{\dfrac{\sqrt{7} }{7}}$

d)

$\dfrac{4}{3\sqrt{3} }=\dfrac{4}{3\sqrt{3} } \times \dfrac{\sqrt{3}}{\sqrt{3}}=\dfrac{4\sqrt{3}}{3\times 3}=\boxed{\dfrac{4\sqrt{3} }{9}}$

e)

$\dfrac{2}{\sqrt{5} }=\dfrac{2}{\sqrt{5} } \times\dfrac{\sqrt{5} }{\sqrt{5}}=\boxed{\dfrac{2\sqrt{5} }{5}}$

f)

$\dfrac{7}{5\sqrt{2}}=\dfrac{7}{5\sqrt{2}} \times\dfrac{\sqrt{2} }{\sqrt{2} } =\dfrac{7\sqrt{2} }{5 \times 2}=\boxed{\dfrac{7\sqrt{2} }{10} }$

Bons Estudos :)