Question

Difficulty: HardProperties of Exponents in Algebraic Expressions

If aa and bb represent positive real numbers, which of the following is an equivalent form of the expression below?

(a1+b1)2ab\frac{(a^{-1} + b^{-1})^{-2}}{ab}
  1. A
    a2+b2ab\frac{a^2 + b^2}{ab}
  2. B
    (a+b)24ab\frac{(a+b)^2}{4ab}
  3. ab(a+b)2\frac{ab}{(a+b)^2}Answer
  4. D
    (a+b)2ab\frac{(a+b)^2}{ab}
  5. E
    1a2b2(a+b)2\frac{1}{a^2 b^2 (a+b)^2}

Answer

The correct answer is the fraction with abab in the numerator and the square of the sum (a+b)(a+b) in the denominator, which is \frac{ab}{(a+b)^2}.
The correct answer is found by first rewriting a1+b1a^{-1} + b^{-1} as 1a+1b\frac{1}{a} + \frac{1}{b}, which simplifies to a+bab\frac{a+b}{ab} using a common denominator. Raising this to the power of 2-2 gives a2b2(a+b)2\frac{a^2 b^2}{(a+b)^2}. Finally, dividing this expression by abab reduces the exponents of aa and bb by 1, resulting in ab(a+b)2\frac{ab}{(a+b)^2}.

Step-by-Step Solution

1
Rewrite the negative exponents in the numerator as reciprocals.
The expression inside the parentheses becomes 1a+1b\frac{1}{a} + \frac{1}{b}.
By the definition of negative exponents, x1=1xx^{-1} = \frac{1}{x}.
2
Find a common denominator to add the fractions inside the parentheses.
1a+1b=bab+aab=a+bab\frac{1}{a} + \frac{1}{b} = \frac{b}{ab} + \frac{a}{ab} = \frac{a+b}{ab}.
To add fractions, they must share a common denominator, which is the product of aa and bb.
3
Apply the negative power of 2-2 to the simplified fraction.
(a+bab)2=(aba+b)2=a2b2(a+b)2\left(\frac{a+b}{ab}\right)^{-2} = \left(\frac{ab}{a+b}\right)^2 = \frac{a^2 b^2}{(a+b)^2}.
An expression raised to a negative exponent is equal to the reciprocal of the expression raised to the positive exponent.
4
Divide the result by the denominator abab.
a2b2(a+b)21ab=ab(a+b)2\frac{a^2 b^2}{(a+b)^2} \cdot \frac{1}{ab} = \frac{ab}{(a+b)^2}.
Dividing by a term is equivalent to multiplying by its reciprocal, and applying the exponent quotient rule simplifies a2b2ab\frac{a^2 b^2}{ab} to abab.

Key Concept

Properties of exponents including negative power rules, quotient rules, and algebraic fraction addition.

Alternative Method

An alternative method is to substitute small positive integer values for aa and bb. Let a=1a = 1 and b=2b = 2. The expression evaluates to (11+21)212=(1+0.5)22=1.522=(3/2)22=4/92=29\frac{(1^{-1} + 2^{-1})^{-2}}{1 \cdot 2} = \frac{(1 + 0.5)^{-2}}{2} = \frac{1.5^{-2}}{2} = \frac{(3/2)^{-2}}{2} = \frac{4/9}{2} = \frac{2}{9}. Evaluating the correct expression with these values yields 12(1+2)2=29\frac{1 \cdot 2}{(1+2)^2} = \frac{2}{9}, which matches.
Estimated Time:2m 0s
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