Bài giải:

$\begin{align}  & \left( 1999\times 1998+1998\times 1997 \right)\times \left( 1+\frac{1}{2}:1\frac{1}{2}-1\frac{1}{3} \right) \\  & =\left( 1999\times 1998+1998\times 1997 \right)\times \left( 1+\frac{1}{2}:\frac{3}{2}-\frac{4}{3} \right) \\ \end{align}$

$\begin{align} & =\left( 1999\times 1998+1998\times 1997 \right)\times \left( 1+\frac{1}{3}-\frac{4}{3} \right) \\ & =\left( 1999\times 1998+1998\times 1997 \right)\times 0 \\  & =0 \\ \end{align}$