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