1/ Để $\sqrt{5x+10}$ có nghĩa thì $5x+10\ge 0$
$↔5x\ge -10\\↔x\ge -2$
Vậy $x\ge -2$ thì $\sqrt{5x+10}$ có nghĩa
3/ ĐK: $x>0$
$P\,=\left(\dfrac{\sqrt{x^3}+1}{\sqrt x+1}+\sqrt x\right):(x+1)\\\quad =\left(\dfrac{(\sqrt x+1)(x-\sqrt x+1)}{\sqrt x+1}+\sqrt x\right):(x+1)\\\quad =[(x-\sqrt x+1)+\sqrt x]:(x+1)\\\quad =(x-\sqrt x+1+\sqrt x):(x+1)\\\quad =(x+1):(x+1)\\\quad =1$
Vậy $P=1$ với $x>0$
