6. Given $x, y, t \in \mathbf{R}_{+}$, let
$$
M=\frac{x}{x^{2}+t}+\frac{y}{y^{2}+t}, N=\frac{4(x+y)}{(x+y)^{2}+4 t} \text {. }
$$

Then among the following 4 propositions:
(1) If $x y=3 t$, then $M=N$;
(2) If $x y>3 t$, then $M \leqslant N$;
(3) If $x y<3 t$, then $M \geqslant N$;
(4) If $x y=3 t, M=N$ does not necessarily always hold, the number of correct propositions is ( ).
(A) 1
(B) 2
(C) 3
(D) 4
6. A.
$$
\begin{array}{l}
M-N=\frac{x}{x^{2}+t}+\frac{y}{y^{2}+t}-\frac{4(x+y)}{(x+y)^{2}+4 t} \\
=\frac{(x+y)(x y+t)}{x^{2} y^{2}+\left(x^{2}+y^{2}\right) t+t^{2}}-\frac{4(x+y)}{(x+y)^{2}+4 t} .
\end{array}
$$

Let $P=\frac{x y+t}{x^{2} y^{2}+\left(x^{2}+y^{2}\right) t+t^{2}}$.
Substituting $x y=3 t$ into the above equation, we get
$$
P=\frac{4}{x^{2}+y^{2}+10 t}=\frac{4}{(x+y)^{2}+4 t} \text {. }
$$

Therefore, when $x y=3 t$, (1) is correct, and (1) is not incorrect.
Let $x=2 \sqrt{t}, y=3 \sqrt{t}$, then $x y>3 t$, and
$$
M=\frac{7}{10 \sqrt{t}}>\frac{20}{29 \sqrt{t}}=N \text {. }
$$

Let $x=2 \sqrt{t}, y=\sqrt{t}$, then $x y<3 t$, and
$$
M=\frac{9}{10 \sqrt{t}}, N=\frac{12}{13 \sqrt{t}} \text {. }
$$

Thus, $M<N$. Therefore, (2), (3), and (4) are all incorrect.