Simplify the following:
3ab−1+2ba−1\dfrac{3a}{b^{-1}} + \dfrac{2b}{a^{-1}}b−13a+a−12b
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Given,
⇒3ab−1+2ba−1=3a1b+2b1a=3ab+2ab=5ab.\Rightarrow \dfrac{3a}{b^{-1}} + \dfrac{2b}{a^{-1}} = \dfrac{3a}{\dfrac{1}{b}} + \dfrac{2b}{\dfrac{1}{a}}\\[1em] = 3ab + 2ab = 5ab.⇒b−13a+a−12b=b13a+a12b=3ab+2ab=5ab.
Hence, 3ab−1+2ba−1\dfrac{3a}{b^{-1}} + \dfrac{2b}{a^{-1}}b−13a+a−12b = 5ab.
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