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Zigoto
@Zigoto
December 2019
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(UNIFEI-MG–2008) Sejam A = raiz quadrada de (x/y), B = raiz cubica de (y^2 / x) e C = raiz sexta de (x/y)
Então, o produto A.B.C é igual a
A) ³Ѵy
B) ³Ѵx
C) ³Ѵx/y
D) ³Ѵxy
edit1: Segue um anexo da questão por inteiro
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Comentários (3)
A= √(x/y)= (X/Y)^1/2=(x/y)^3/6
B=∛(y^2/x)= (Y^2/X)^1/3=(y^2/x)^2/6
C=raiz-6(x/y)= (x/y)^1/6
ABC=(x/y)^3/6*(y^2/x)^2/6* (x/y)^1/6
ABC=[(X/Y)^3*(Y^2/X)^2*(X/Y)]^(1/6)
ABC=[ X^3/Y^3*Y^4/X^2*X/Y]^(1/6)
ABC=[X^3*Y^-3*Y^4*X^-2*X*Y^-1]^1/6
ABC=[X^(3-2+1)*Y^(-3+4-1)]^(1/6)
ABC=[X^2*Y^0]^1/6)=(X^2)^1/6
= X^2/6=X^(1/3)=∛x
espero de ajudado :)
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B=∛(y^2/x)= (Y^2/X)^1/3=(y^2/x)^2/6
C=raiz-6(x/y)= (x/y)^1/6
ABC=(x/y)^3/6*(y^2/x)^2/6* (x/y)^1/6
ABC=[(X/Y)^3*(Y^2/X)^2*(X/Y)]^(1/6)
ABC=[ X^3/Y^3*Y^4/X^2*X/Y]^(1/6)
ABC=[X^3*Y^-3*Y^4*X^-2*X*Y^-1]^1/6
ABC=[X^(3-2+1)*Y^(-3+4-1)]^(1/6)
ABC=[X^2*Y^0]^1/6)=(X^2)^1/6 = X^2/6=X^(1/3)=∛x
espero de ajudado :)