Perturbation of the energy-momentum tensor: mistake in my computations or in the book? In the book Cosmology - S. Weinberg eqauation $(5.1.28)$ is
$$
\delta T^{\mu}_{\;\;\nu} = \bar{g}^{\mu\lambda} [\delta T_{\lambda\nu} - h_{\lambda\kappa} \bar{T}^{\kappa}_{\;\;\nu}  ] \tag{5.1.28}
$$
In which $\bar{g}$ is the unperturbed metric and $h$ is the first order perturbation to the metric defined by
$$
g_{\mu\nu} = \bar{g}_{\mu\nu} + h_{\mu\nu}
$$
and
$$
h^{\mu\nu} = - \bar{g}^{\mu\rho}\bar{g}^{\nu\sigma}h_{\rho\sigma}
$$
$\bar{T}$ is the unperturbed energy-momentum tensor, which in this case take the form of a perfect fluid
$$
T_{\mu\nu} = pg_{\mu\nu} + (p+\rho)u_\mu u_\nu
$$
At this point $h$ is completely general and there is no need to take a gauge where $h_{00}$, in fact he even writes $h_{00} = -E$.
He then writes $\delta T_{00} = -\bar{\rho} h_{00} +  \delta \rho$ and so far I'm ok with it. My problem is with equation $(5.1.43)$ in which he writes:

The perturbed mixed components (5.1.28)
of the energy-momentum tensor, which are needed in the conservation laws,
now take the form

$$
\delta T^0_{\;\;0} = -\delta \rho.
$$
But from equation $(5.1.28)$, shouldn't this be
$$
\delta T^0_{\;\;0} = -\delta \rho - \bar{g}^{00} h_{00} \bar{T}^{0}_{\;\;0}
$$
?
can you please explain to me why that second term I've written disappear?
I have checked the errata corrige and that's not a typo, I happen to make quite a lot of mistakes doing these kind of computations so my bet is that I'm doing something wrong.
 A: Let's go slowly using the fact that $\bar{g}^{0\lambda}$ and $\bar{g}_{0\lambda}$ are only nonzero for $\lambda = 0$.
\begin{align}
\delta T^0_{\;\;0} &= \bar{g}^{00} \left [ \delta T_{00} - h_{0\kappa} \bar{T}^\kappa_{\;\;0} \right ] \\
&= -\left [ \delta T_{00} - h_{0\kappa} \bar{T}^\kappa_{\;\;0} \right ] \\
&= -\left [ -\bar{\rho} h_{00} + \delta \rho - h_{0\kappa} \bar{T}^\kappa_{\;\;0} \right ] \\
&= -\left [ -\bar{\rho} h_{00} + \delta \rho - h_{0\kappa} \bar{g}^{\kappa \lambda} \bar{T}_{\lambda 0} \right ]
\end{align}
If this makes sense so far, there is only one more thing we need to use: the fact that the frame is comoving with the perfect fluid. This wasn't explicitly stated in your question but it's the only thing that makes sense given the context. So I would make sure Weinberg mentions it. This would mean that, much like $p$ and $\rho$ are expanded around unperturbed values $\bar{p}$ and $\bar{\rho}$,
\begin{equation}
u_\mu = \bar{u}_\mu + \delta u_\mu = (1, 0, 0, 0) + \delta u_\mu.
\end{equation}
This leads to a diagonal
\begin{equation}
\bar{T}_{\mu\nu} = \mathrm{diag}(\bar{\rho}, \bar{p}, \bar{p}, \bar{p}).
\end{equation}
Plugging this into the expression on top, $\bar{T}$ singles out $\lambda = 0$ which means $\bar{g}$ singles out $\kappa = 0$ and then we have
\begin{align}
\delta T^0_{\;\;0} &= -\left [ -\bar{\rho} h_{00} + \delta \rho - h_{0\kappa} \bar{g}^{\kappa \lambda} \bar{T}_{\lambda 0} \right ] \\
&= -\left [ -\bar{\rho} h_{00} + \delta \rho + h_{0 0} \bar{T}_{0 0} \right ] \\
&= -\delta \rho
\end{align}
