Question:

<div class="solution"> <p><strong>Step 1:</strong> The problem asks us to find the coefficient of \(x^{-5}\) in the binomial expansion of \(\left(\dfrac{x+1}{x^{2/3}-x^{1/3}+1}-\dfrac{x-1}{x-x^{1/2}}\right)^{10}\), where \(x\neq 0,1\). To start, we should simplify the given expression to make it easier to work with.</p> <p><strong>Step 2:</strong> We simplify the given expression: \[ \dfrac{x+1}{x^{2/3}-x^{1/3}+1} = \dfrac{x+1}{(x^{1/3}-1)(x^{1/3}+1)+1} = \dfrac{x+1}{x^{1/3}(x^{1/3}-1)+x^{1/3}+2} = \dfrac{x^{2/3}(x^{1/3}+1)}{x^{2/3}-x^{1/3}+1} = x^{1/3} \] and \[ \dfrac{x-1}{x-x^{1/2}} = \dfrac{x-1}{x^{1/2}(x^{1/2}-1)} = \dfrac{x^{1/2}(x^{1/2}-1)}{x^{1/2}(x^{1/2}-1)} = x^{1/2}. \] Therefore, the given expression simplifies to \((x^{1/3}-x^{1/2})^{10}\).</p> <p><strong>Step 3:</strong> The term that will produce \(x^{-5}\) in the expansion of \((x^{1/3}-x^{1/2})^{10}\) must have the form \(x^{10k/3} \cdot x^{-10(1-k)/2} = x^{-5}\), where \(k\) is an integer between 0 and 10 (inclusive). This leads to the equation \(\dfrac{10k}{3} - \dfrac{10(1-k)}{2} = -5\). Solving for \(k\), we get \(\dfrac{10k}{3} - 5 + \dfrac{10k}{2} = -5\), which simplifies to \(\dfrac{20k+30k}{6} = 0\), and further to \(50k = 0\). This implies \(k = 0\), but since we are looking for the term that gives \(x^{-5}\), we should consider the term where the powers of \(x\) from both factors combine to give \(x^{-5}\). This corresponds to the term where \(x^{1/3}\) is raised to the power of \(k\) and \(x^{1/2}\) to the power of \(10-k\), such that their exponents sum to \(-5\). The correct approach involves finding the term in the expansion that matches \(x^{-5}\), which can be achieved by considering the general term in the binomial expansion: \(\binom{10}{k} (x^{1/3})^k (-x^{1/2})^{10-k}\). We seek the term where the exponent of \(x\) is \(-5\), so \(\dfrac{k}{3} - \dfrac{10-k}{2} = -5\). Solving this equation for \(k\), we multiply through by 6 to clear the fractions: \(2k - 3(10-k) = -30\), which simplifies to \(2k - 30 + 3k = -30\), and further to \(5k = 0\), leading to \(k = 0\). However, this calculation seems to have been misdirected, as we need to correctly identify the term that contributes to \(x^{-5}\). Let's correct the approach by directly evaluating the term that would give us \(x^{-5}\) from the expansion of \((x^{1/3}-x^{1/2})^{10}\). The correct term to consider should satisfy the condition that its exponent equals \(-5\), and since we are dealing with a binomial expansion, we should apply the binomial theorem correctly to identify this term.</p> <p><strong>Step 4:</strong> Applying the binomial theorem, the general term in the expansion of \((x^{1/3}-x^{1/2})^{10}\) is given by \(\binom{10}{k} (x