a)Ta có $4^2=16\Rightarrow \sqrt{15}<4\\\Rightarrow 12+\sqrt{15}<16\Rightarrow \sqrt{12+\sqrt{15}}<4$
Tương tự như vậy suy ra
$\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{15}}}}<4$
b)Ta có $\sqrt{15}<4;\ \sqrt{24}<5\\\Rightarrow \sqrt{15}+\sqrt{24}<9$
$1<\sqrt{2}<1,5;\ 2<\sqrt{5}<2,5\\\Rightarrow 3<\sqrt{2}+\sqrt{5}<4$
$\Rightarrow \sqrt{2}+\sqrt{5}+\sqrt{15}+\sqrt{24}<13$