Alguém me ajuda? racionalização das seguintes expressões

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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Alguém me ajuda? racionalização das seguintes expressões

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