8. Solve the equation $\frac{x^{4}}{2 x+1}+x^{2}=6(2 x+1)$.

Answer. $x=-3 \pm \sqrt{6}, x=2 \pm \sqrt{6}$.
By dividing both sides of the equation by $(2 x+1)$, we get $\left(\frac{x^{2}}{2 x+1}\right)^{2}+\frac{x^{2}}{2 x+1}-6=0$, from which $\frac{x^{2}}{2 x+1}=2$ or $\frac{x^{2}}{2 x+1}=-3$. In the first case, we get $x^{2}-4 x-2=0, x=2 \pm \sqrt{6}$, and in the second case $x^{2}+6 x+3=0, x=-3 \pm \sqrt{6}$.

1 Simplify the fraction $\frac{5 x+y-2 \sqrt{5 x y}}{\sqrt{-5 x}+\sqrt{-y}}$.

Answer. $-\sqrt{-5 x}-\sqrt{-y}$.

Solution Let $-x=u,-y=v$, where $u \geq 0, v \geq 0$. Then the given fraction equals $\frac{-5 u-v-2 \sqrt{5 u v}}{\sqrt{5 u}+\sqrt{v}}=\frac{-(\sqrt{5 u}+\sqrt{v})^{2}}{\sqrt{5 u}+\sqrt{v}}=-(\sqrt{5 u}+\sqrt{v})=-\sqrt{-5 x}-\sqrt{-y}$.