Question

alguém me ajuda? racionalização das seguintes expressões

Answer 1

a) $\dfrac{4}{\sqrt{2}-6}=\dfrac{4\cdot(\sqrt{2}+6)}{(\sqrt{2}-6})(\sqrt{2}+6)}=\dfrac{4\sqrt{2}+24}{(\sqrt{2})^2-6^2}$

$=\dfrac{4\sqrt{2}+24}{2-36}=\dfrac{4\sqrt{2}+24}{-34}=\dfrac{`-2\sqrt{2}-12}{17}$

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

$=\dfrac{10+2\sqrt{3}}{25-3}=\dfrac{10+2\sqrt{3}}{22}=\dfrac{5+\sqrt{3}}{11}$

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

$=\dfrac{7\sqrt{3}+7\sqrt{2}}{3-2}=7\sqrt{3}+7\sqrt{2}$

d) $\dfrac{2\sqrt{3}}{\sqrt{7}+6}=\dfrac{2\sqrt{3}\cdot(\sqrt{7}-6)}{(\sqrt{7}+6)(\sqrt{7}-6)}=\dfrac{2\sqrt{21}-12\sqrt{3}}{(\sqrt{7})^2-6^2}$

$=\dfrac{2\sqrt{21}-12\sqrt{3}}{7-36}=\dfrac{2\sqrt{21}-12\sqrt{3}}{-29}=\dfrac{-2\sqrt{21}+12\sqrt{3}}{29}$

e) $\dfrac{12}{9+\sqrt{7}}=\dfrac{12\cdot(9-\sqrt{7})}{(9+\sqrt{7})(9-\sqrt{7})}=\dfrac{108-12\sqrt{7}}{9^2-(\sqrt{7})^2}$

$=\dfrac{108-12\sqrt{7}}{81-7}=\dfrac{108-12\sqrt{7}}{74}=\dfrac{54-6\sqrt{7}}{37}$

f) $\dfrac{\sqrt{5}}{\sqrt{5}+\sqrt{3}}=\dfrac{\sqrt{5}\cdot(\sqrt{5}-\sqrt{3})}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}$

$=\dfrac{5-\sqrt{15}}{(\sqrt{5})^2-\sqrt{3})^2}=\dfrac{5-\sqrt{15}}{5-3}=\dfrac{5-\sqrt{15}}{2}$

g) $\dfrac{4\sqrt{3}}{3+\sqrt{7}}=\dfrac{4\sqrt{3}\cdot(3-\sqrt{7})}{(3+\sqrt{7})(3-\sqrt{7})}=\dfrac{12\sqrt{3}-4\sqrt{21}}{3^2-(\sqrt{7})^2}$

$=\dfrac{12\sqrt{3}-4\sqrt{21}}{9-7}=\dfrac{12\sqrt{3}-4\sqrt{21}}{2}=6\sqrt{3}-2\sqrt{21}$

h) $\dfrac{\sqrt{5}}{6-\sqrt{2}}=\dfrac{\sqrt{5}\cdot(6+\sqrt{2})}{(6-\sqrt{2})(6+\sqrt{2})}=\dfrac{6\sqrt{5}+\sqrt{10}}{6^2-(\sqrt{2})^2}$

$=\dfrac{6\sqrt{5}+\sqrt{10}}{36-2}=\dfrac{6\sqrt{5}+\sqrt{10}}{34}$

i) $\dfrac{4\sqrt{3}}{\sqrt{5}+\sqrt{2}}=\dfrac{4\sqrt{3}\cdot(\sqrt{5}-\sqrt{2})}{(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})}$

$=\dfrac{4\sqrt{15}-4\sqrt{6}}{(\sqrt{5})^2-(\sqrt{2})^2}=\dfrac{4\sqrt{15}-4\sqrt{6}}{5-2}=\dfrac{4\sqrt{15}-4\sqrt{6}}{3}$

j) $\dfrac{9}{\sqrt{3}+3}=\dfrac{9\cdot(\sqrt{3}-3)}{(\sqrt{3}+3)(\sqrt{3}-3)}=\dfrac{9\sqrt{3}-27}{(\sqrt{3})^2-3^2}$

$=\dfrac{9\sqrt{3}-27}{3-9}=\dfrac{9\sqrt{3}-27}{-6}=\dfrac{-3\sqrt{3}+9}{2}$

k) $\dfrac{\sqrt{3}}{\sqrt{7}+\sqrt{3}}=\dfrac{\sqrt{3}\cdot(\sqrt{7}-\sqrt{3})}{(\sqrt{7}+\sqrt{3})(\sqrt{7}-\sqrt{3})}$

$=\dfrac{\sqrt{21}-3}{(\sqrt{7})^2-(\sqrt{3})^2}=\dfrac{\sqrt{21}-3}{7-3}=\dfrac{\sqrt{21}-3}{4}$

l) $\dfrac{6\sqrt{2}}{11-\sqrt{3}}=\dfrac{6\sqrt{2}\cdot(11+\sqrt{3})}{(11+\sqrt{3})(11-\sqrt{3})}=\dfrac{66\sqrt{2}+6\sqrt{6}}{11^2-(\sqrt{3})^2}$

$=\dfrac{66\sqrt{2}+6\sqrt{6}}{121-3}=\dfrac{66\sqrt{2}+6\sqrt{6}}{118}=\dfrac{33\sqrt{2}+3\sqrt{6}}{59}$

m) $\dfrac{6}{\sqrt{5}+13}=\dfrac{6\cdot(\sqrt{5}-13)}{(\sqrt{5}+13)(\sqrt{5}-13)}=\dfrac{6\sqrt{5}-78}{(\sqrt{5})^2-13^2}$

$=\dfrac{6\sqrt{5}-78}{5-169}=\dfrac{6\sqrt{5}-78}{-164}=\dfrac{-3\sqrt{5}+39}{82}$

n) $\dfrac{8}{4-\sqrt{10}}=\dfrac{8\cdot(4+\sqrt{10})}{(4-\sqrt{10})(4+\sqrt{10})}=\dfrac{32+8\sqrt{10}}{4^2-(\sqrt{10})^2}$

$=\dfrac{32+8\sqrt{10}}{16-10}=\dfrac{32+8\sqrt{10}}{6}=\dfrac{16+4\sqrt{10}}{3}$

Answer 2

Resposta:

ajudou mt obrigado

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