Question

aYudenme pofavor no me ignoren es para mañanasi no raspare la materia alguien que lo sepa resolver

Answer 1

$t = \frac{k - {d}^{2} + \frac{4\pi}{ \sqrt{4f} } }{4kx + 2\pi - d} \\ t(4kx + 2\pi - d) = k - {d}^{2} + \frac{4\pi}{ \sqrt{4f} } \\ t(4kx + 2\pi - d) - k + {d}^{2} = \frac{4\pi}{ \sqrt{4f} } \\ \sqrt{4f} = \frac{4\pi}{ t(4kx + 2\pi - d) - k + {d}^{2} } \\2 \sqrt{f} =\frac{4\pi}{ t(4kx + 2\pi - d) - k + {d}^{2} } \\ \sqrt{f} = \frac{4\pi}{2(t(4kx + 2\pi - d) - k + {d}^{2}) } \\ f = {(\frac{4\pi}{2(t(4kx + 2\pi - d) - k + {d}^{2})})}^{2} \\ f = {(\frac{2\pi}{t(4kx + 2\pi - d) - k + {d}^{2}})}^{2}$