Câu 1: $(x+4)^{2} - 6\sqrt{x^{3}+3x}=13$ (ĐKXĐ: x>0)$<=>(x+4)^{2}-13 = 6\sqrt{x^{3}+3x}$
$<=>(x+4)^4 -26(x+4)^2+169=36(x^3+3x)$
$<=>(x^4+16x^3+96x^2+256x+256)-(26x^2+208x+416)+169=36x^3+108x$
$<=>x^4-20x^3+70x^2-60x+9=0$
$<=>(x-1)(x-3)(x^2-16x+3)=0$
$<=> x=1$ hoặc $x=3$ hoặc $x=8\pm \sqrt{61} $ (TMĐK)
Câu 2: $2\sqrt[3]{x-2}+5\sqrt{x+1}-12=0$ (ĐKXĐ: x > -1)
$<=>(2\sqrt[3]{x-2}-2)+(5\sqrt{x+1}-10)=0$
$<=>\frac{8(x-2)-8}{4\sqrt[3]{(x-2)^2}+4\sqrt[3]{x-2}+4}+\frac{5(x+1-4)}{\sqrt{x+1} +2}=0$
$<=>\frac{8(x-3)}{4\sqrt[3]{(x-2)^2}+4\sqrt[3]{x-2}+4}+\frac{5(x-3)}{\sqrt{x+1}+2}=0$
$<=>(x-3)(\frac{8}{4\sqrt[3]{(x-2)^2}+4\sqrt[3]{x-2}+4}+\frac{5}{\sqrt{x+1}+2})=0$ (*)
Vì $\frac{8}{4\sqrt[3]{(x-2)^2}+4\sqrt[3]{x-2}+4}+\frac{5}{\sqrt{x+1}+2}>0$ Với mọi x > -1
Nên (*) <=> $x-3=0$ $<=> x=3$