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Example 1
$(3a - \dfrac{1}{3a})^2$ is equivalent to
- $9a^2+\dfrac{1}{9a^2}$
- $9a^2-\dfrac{1}{9a^2}$
- $9a^2-2+\dfrac{1}{9a^2}$
- $9a^2+2+\dfrac{1}{9a^2}$
- 1
Solution
- The given expression is the square of a difference: $(x-y)^2=x^2-2xy+y^2$
Apply it to the given expression with $x = 3a$ and $y=\dfrac{1}{3a}$
$(3a - \dfrac{1}{3a})^2=(3a)^2-2(3a)(\dfrac{1}{3a})+(\dfrac{1}{3a})^2$
- Simplify the middle term as follows:
$2(3a)(\dfrac{1}{3a}) = 2\dfrac{3a}{3a}=2\dfrac{\colorcancel{red}{3a}}{\colorcancel{red}{3a}}=2$
Hence
$(3a - \dfrac{1}{3a})^2=9a^2-2+\dfrac{1}{9a^2}$
Answer C
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