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\[\frac{1}{(a + b)^\frac{1}{2}} = (a + b)^{\frac{1}{2}}\]
If the above is true, which of the following must be true?
\[(A) \quad a = 0\]
\[(B) \quad \sqrt{a + b} =  1\]
\[(C) \quad \sqrt{a + b} = 0\]
\[(D) \quad a + b = 1\]
\[(E) \quad (a + b)^2 = 0\]
Strategic advice:
Here is an expression that is equal to the reciprocal of itself. Can you think of numbers that have those properties?
Answer: (D)
Don't let the fractions in the exponents throw you off. The first step of such problems is to crossmultiply.
\[\frac{1}{(a + b)^\frac{1}{2}} = (a + b)^{\frac{1}{2}}\]
\[1 = \frac{1}{(a + b)^\frac{1}{2}} \quad * \quad (a + b)^{\frac{1}{2}}\]
\[(a+b)^{1} = {\frac{1}{a + b}}\]
Since this fraction is equal to 1, we know the reciprocal is also equal to 1.
1 = a + b, or choice (D).
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