Correct Answer: 2

Solution : Given: $x=\frac{8ab}{a+b}(a\neq b)$
Then, putting the value of $x$ in expression $\frac{x+4a}{x–4a}+\frac{x+4b}{x–4b}$, we have,
= $\frac{\frac{8ab}{a+b}+4a}{\frac{8ab}{a+b}–4a}+\frac{\frac{8ab}{a+b}+4b}{\frac{8ab}{a+b}–4b}$
= $\frac{\frac{8ab+4a^{2}+4ab}{a+b}}{\frac{8ab–4a^{2}–4ab}{a+b}}+\frac{\frac{8ab+4ab+4b^{2}}{a+b}}{\frac{8ab–4ab–4b^{2}}{a+b}}$
= $\frac{{8ab+4a^{2}+4ab}}{8ab–4a^{2}–4ab}+\frac{8ab+4ab+4b^{2}}{8ab–4ab–4b^{2}}$
= $\frac{{12ab+4a^{2}}}{4ab–4a^{2}}+\frac{12ab+4b^{2}}{4ab–4b^{2}}$
= $\frac{{4a(3b+a)}}{4a(b–a)}+\frac{4b(3a+b)}{4b(a–b)}$
= $\frac{{(3b+a)}}{(b–a)}+\frac{(3a+b)}{(a–b)}$
= $\frac{{(3b+a–3a–b)}}{(b–a)}$
= $\frac{{(2b–2a)}}{(b–a)}$
= $\frac{{2(b–a)}}{(b–a)}$
= 2
Hence, the correct answer is 2.