Question

como fazer as raízes quadradas não exatas que tem um número do lado de fora?​

Answer 1

É preciso fatorar o numero dentro da raiz em fatores primos para podermos simplificar as raízes, acompanhe:

a)

$\sqrt{75}=\sqrt{3~.~5~.~5}=\sqrt{3~.~5^2}=\boxed{5\sqrt{3}}\\\\\sqrt{27}=\sqrt{3~.~3~.~3}=\sqrt{3~.~3^2}=\boxed{3\sqrt{3}}\\\\\\\sqrt{75}+\sqrt{27}=5\sqrt{3}+3\sqrt{3}=\boxed{8\sqrt{3}}$

b)

$\sqrt{98}-\sqrt{32}=\sqrt{2~.~7~.~7}-\sqrt{2~.~2~.~2~.~2~.~2}\\\\\sqrt{98}-\sqrt{32}=\sqrt{2~.~7^2}-\sqrt{2~.~2^2~.~2^2}\\\\\sqrt{98}-\sqrt{32}=7\sqrt{2}-4\sqrt{2}\\\\\boxed{\sqrt{98}-\sqrt{32}=3\sqrt{2}}$

c)

$2\sqrt{40}+3\sqrt{10}=2\sqrt{2~.~2~.~10}+3\sqrt{10}\\\\2\sqrt{40}+3\sqrt{10}=2\sqrt{2^2~.~10}+3\sqrt{10}\\\\2\sqrt{40}+3\sqrt{10}=2.2\sqrt{10}+3\sqrt{10}\\\\\boxed{2\sqrt{40}+3\sqrt{10}=7\sqrt{10}}$

d)

$5\sqrt{45}-10\sqrt{20}=5\sqrt{3~.~3~.~5}-10\sqrt{2~.~2~.~5}\\\\5\sqrt{45}-10\sqrt{20}=5\sqrt{3^2~.~5}-10\sqrt{2^2~.~5}\\\\5\sqrt{45}-10\sqrt{20}=5.3\sqrt{5}-10.2\sqrt{5}\\\\\boxed{5\sqrt{45}-10\sqrt{20}=-5\sqrt{5}}$

e)

$5\sqrt{12}-4\sqrt{27}=5\sqrt{2~.~2~.~3}-4\sqrt{3~.~3~.~3}\\\\5\sqrt{12}-4\sqrt{27}=5\sqrt{2^2~.~3}-4\sqrt{3^2~.~3}\\\\5\sqrt{12}-4\sqrt{27}=5.2\sqrt{3}-4.3\sqrt{3}\\\\\boxed{5\sqrt{12}-4\sqrt{27}=-2\sqrt{3}}$

f)

$3\sqrt[3]{16}+\sqrt[3]{54}=3\sqrt[3]{2~.~2~.~2~.~2}+\sqrt[3]{2~.~3~.~3~.~3}\\\\3\sqrt[3]{16}+\sqrt[3]{54}=3\sqrt[3]{2^3~.~2}+\sqrt[3]{2~.~3^3}\\\\3\sqrt[3]{16}+\sqrt[3]{54}=3.2\sqrt[3]{2}+2\sqrt[3]{2}\\\\\boxed{3\sqrt[3]{16}+\sqrt[3]{54}=8\sqrt[3]{2}}$