How was this approximation made? (Archimedes law and weights)

Question

Excerpt from textbook:

According to Archimedes' law the weight of a body of mass $m$ and density $\rho$ inside air is: $$G=mg\left(1 - \frac{\rho_v}{\rho}\right)$$ Where $\rho_v$ is the density of air.
A "weight" of mass $m_t$ and density $\rho_t$ in air weights $$G_t = m_tg\left(1-\frac{\rho_v}{\rho_t}\right)$$ If using a balance scale one determined $G=G_t$ one would get: $$m=m_t\frac{(1 - \rho_v/\rho_t)}{(1-\rho_v/\rho)}$$ i.e. The mass of the weight and the mass of the body are not equal. In real measurments $\rho_v/\rho_t\ll1$ and $\rho_v/\rho\ll1$ so the previous equation can be approximately written as: $$m=m_t\left(1+\frac{\rho_v}{\rho} - \frac{\rho_v}{\rho_t}\right)$$

How was the last approximation made? I tried deriving the 4th equation starting from the 3rd equation:

$$m- m_t - m\frac{\rho_v}{\rho} + m_t\frac{\rho_v}{\rho_t} = 0$$ Using approximation magic (perhaps the fact $\rho_v/\rho_t\ll1$ and $\rho_v/\rho\ll1$) I may turn $m$ into $m_t$ in the 3rd addend and factor $m_t$ : $$m- m_t - m_t\frac{\rho_v}{\rho} + m_t\frac{\rho_v}{\rho_t} = 0$$ $$m- m_t\left(1 + \frac{\rho_v}{\rho} - \frac{\rho_v}{\rho_t}\right) = 0$$ That gives me the formula I wanted.
But it doesn't make any sense and probably another line of logic was used.

Answer 1

For x<<1 the following common approximation is true:

1/(1-x)=1+x

So for y<<1 we get:

(1-y)/(1-x)=(1-y)(1+x)=1-y+x

The -xy component in the last equation is negligible, as it contains a product of two small values.