C-04-Q094mediumsingle_mcqX=(2a−1+3b−1)−1X = (2a^{-1} + 3b^{-1})^{-1}X=(2a−1+3b−1)−1. Then X=?X = ?X=?aab2b+3a\dfrac{ab}{2b + 3a}2b+3aabb2a+3b2a + 3b2a+3bca+bab\dfrac{a + b}{ab}aba+bd2a+3b\dfrac{2}{a} + \dfrac{3}{b}a2+b3ব্যাখ্যাThe inner sum is 2a+3b=2b+3aab\frac{2}{a}+\frac{3}{b}=\frac{2b+3a}{ab}a2+b3=ab2b+3a; taking the −1-1−1 power inverts it to ab2b+3a\frac{ab}{2b+3a}2b+3aab.