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Question

Câu 1:

$(x+4)^{2} - 6\sqrt{x^{3}+3x}=13$

Câu 2:

$2\sqrt[3]{x-2}+5\sqrt{x+1}-12=0$

hothinhtls · Answer 1 · 6/17/2014 9:26:43 AM

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 (*) <=>

$x-3=0$

$<=> x=3$

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