B) 2t-1 12 5t', 7n+2kn г) 9n-2k + n 2k

в) $$\frac{1}{2t-1} - \frac{2}{5t} = \frac{1 \cdot 5t - 2 \cdot (2t-1)}{5t(2t-1)} = \frac{5t - 4t + 2}{5t(2t-1)} = \frac{t+2}{5t(2t-1)}$$ г) $$\frac{7n+2k}{9n-2k} + \frac{n}{2k} = \frac{(7n+2k) \cdot 2k + n \cdot (9n-2k)}{2k(9n-2k)} = \frac{14nk + 4k^2 + 9n^2 - 2nk}{2k(9n-2k)} = \frac{12nk + 4k^2 + 9n^2}{2k(9n-2k)}$$ Ответ: в) $$\frac{t+2}{5t(2t-1)}$$; г) $$\frac{12nk + 4k^2 + 9n^2}{2k(9n-2k)}$$