Question

√7 sobre √2 = prfv me ajuda

Answer 1

Resposta:

$\dfrac{\sqrt{14}}{2}$

Explicação passo-a-passo:

São raízes inexatas e não temos o valor númerico (exceto caso você tenha esquecido de colocar), por isso só podemos racionalizar o denominador (tirar a raiz de baixo):

$\dfrac{\sqrt{7}}{\sqrt{2}}=\\\dfrac{\sqrt{7}}{\sqrt{2}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}}= \dfrac{\sqrt{14}}{2}$

Questões dos comentários:

$\dfrac{6}{\sqrt{2}+1}=\\\dfrac{6}{\sqrt{2}+1}\cdot\dfrac{\sqrt{2}-1}{\sqrt{2}-1}=\dfrac{6\sqrt{2}-6}{2-1}=\dfrac{6\sqrt{2}-6}{1} =6\sqrt{2}-6$

$\dfrac{\sqrt{3}}{3-\sqrt{3}}=\\\dfrac{\sqrt{3}}{3-\sqrt{3}}=\dfrac{\sqrt{3}}{3-\sqrt{3}}\cdot\dfrac{3+\sqrt{3}}{3+\sqrt{3}}=\dfrac{3\sqrt{3}+3}{9-3}=\dfrac{3\sqrt{3}+3}{6}$

Simplificando:

$\dfrac{\sqrt{3}+1}{2}$

$\dfrac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}=\\\dfrac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}=\dfrac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}\cdot\dfrac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}=\dfrac{(\sqrt{5}-\sqrt{2})^{2}}{5-2}=\dfrac{(\sqrt{5}-\sqrt{2})^{2}}{3}=\\\\\\\dfrac{5-2\sqrt{5}\cdot\sqrt{2}+2}{3}= \dfrac{7-2\sqrt{10}}{3}$

Mais algumas outras questões:

$a)\dfrac{3}{2}+\dfrac{9}{4}=\dfrac{3}{4}\cdot2+\dfrac{9}{4}=\dfrac{6}{4}+\dfrac{9}{4}=\dfrac{15}{4}\\mmc(2, 4)=4\\\\b)\dfrac{5}{2}-\dfrac{12}{5}=\dfrac{5}{10}\cdot5-\dfrac{12}{10}\cdot2=\dfrac{25}{10}-\dfrac{24}{10}=\dfrac{1}{10}\\mmc(2, 5)=10\\\\c) \dfrac{28}{3}+\dfrac{2-7}{4}=\dfrac{28}{12}\cdot4+\dfrac{-5}{12}\cdot3=\dfrac{112-15}{12}=\dfrac{97}{12}\\mmc(3, 4)=12\\\\d) \dfrac{7}{2}\cdot4=\dfrac{28}{2}=14\\\\e)9\cdot\dfrac{5}{7}=\dfrac{45}{7}$

$f)\dfrac{5}{4}\cdot\dfrac{8}{7}=\dfrac{40}{28}=\dfrac{10}{7}\\\\g) Respondida.\\\\h)\dfrac{10}{3}:9=\dfrac{10}{9\cdot3}=\dfrac{10}{27}\\\\i)8:\dfrac{7}{3}=8\cdot\dfrac{3}{7}=\dfrac{24}{7}$