1. Given real numbers $a>0, b>0$, satisfying $a+\sqrt{a}=2008, b^{2}+b=2008$. Then the value of $a+b$ is $\qquad$
$=, 1.2008$
For the convenience of calculation, let $2008=m$.
From $a+\sqrt{a}-m=0$, we get
$$
\sqrt{a}=\frac{-1+\sqrt{1+4 m}}{2} \text{, }
$$

which means $a=\frac{1+2 m-\sqrt{1+4 m}}{2}$.
From $b^{2}+b-m=0$, we get
$$
b=\frac{-1+\sqrt{1+4 m}}{2} \text{. }
$$

Therefore, $a+b=m=2008$.