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How to simplify Sqrt[1/x] Sqrt[x]?

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In my expression, there appear terms of the form A^(B Sqrt[1/C] Sqrt[C]). Mathematica doesn't realize that this is just simply A^B. I tried telling it explicitly by some replacement rule. This works for simple cases, but somehow if the form above is embedded in a larger expression it does not do this replacement rule.

For example,

(9 + 9 E^((4 t Sqrt[Λ])/Sqrt[3]) -  12 y^2 Λ - 12 z^2 Λ+ 6 y^2 Sqrt[1/Λ] Λ^(3/2) + 6 z^2 Sqrt[1/Λ] Λ^(3/2) + x^4 Λ^2 + y^4 Λ^2 + 2 y^2 z^2 Λ^2 + z^4 Λ^2 + 2 x^2 Λ (-6 +    3 Sqrt[1/Λ]      Sqrt[Λ] + (y^2 + z^2) Λ) + 6 E^((2 t Sqrt[Λ])/Sqrt[  3]) (3 - 2 z^2 Λ+    x^2 (-2 +       Sqrt[1/Λ]         Sqrt[Λ]) Λ+    y^2 (-2 +       Sqrt[1/Λ]         Sqrt[Λ]) Λ+    z^2 Sqrt[1/Λ] Λ^(3/2)))/(9 + 9 E^((4 t Sqrt[Λ])/Sqrt[3]) -  12 y^2 Λ - 12 z^2 Λ+ 6 y^2 Sqrt[1/Λ] Λ^(3/2) + 6 z^2 Sqrt[1/Λ] Λ^(3/2) + x^4 Λ^2 + y^4 Λ^2 + 2 y^2 z^2 Λ^2 + z^4 Λ^2 + 2 x^2 Λ (-6 +    3 Sqrt[1/Λ]      Sqrt[Λ] + (y^2 + z^2) Λ) -  6 E^((2 t Sqrt[Λ])/Sqrt[  3]) (-3 + x^2 Sqrt[1/Λ] Λ^(3/2) +    y^2 Sqrt[1/Λ] Λ^(3/2) +    z^2 Sqrt[1/Λ] Λ^(3/2)))/. A_^(B_. Sqrt[1/Λ] Sqrt[Λ]) :> A^B

Could anyone help me?


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