6. Let $x, y \in \mathbf{R}^{+}$, and $\frac{19}{x}+\frac{98}{y}=1$, then the minimum value of $x+y$ is
\begin{array}{l}\text { 6. Let } \sin ^{2} \alpha=\frac{19}{x}, \cos ^{2} \alpha=\frac{98}{y}, \therefore x=\frac{19}{\sin ^{2} \alpha}, y=\frac{98}{\cos ^{2} \alpha}, \\ \therefore x+y=\frac{19}{\sin ^{2} \alpha}+\frac{98}{\cos ^{2} \alpha}=19 \csc ^{2} \alpha+98 \sec ^{2} \alpha=19\left(1+\cot ^{2} \alpha\right)+98\left(1+\tan ^{2} \alpha\right)=19 \cot ^{2} \alpha+98 \tan ^{2} \alpha+ \\ 19+98 \geqslant 2 \sqrt{19 \times 98}+117+14 \sqrt{38} .\end{array}