Chủ đề này có 1 trả lời

[trang2004]

$x^2+3x\sqrt[3]{3x+2}-12+\frac{1}{\sqrt{x}}=\frac{\sqrt{x}+8}{x}$

[Frosty Flame]

$x^2+3x\sqrt[3]{3x+2}-12+\frac{1}{\sqrt{x}}=\frac{\sqrt{x}+8}{x}$

ĐKXĐ: $x>0$

Pt đã cho

$x^3+3x^2\sqrt[3]{3x+2}-12x-8=0<=>4(x^3-3x-2)+3x^2(x-\sqrt[3]{3x+2})=0$

$<=>(x^3-3x-2)[4+\frac{3x^2}{\sqrt[3]{(3x+2)^2}+x\sqrt[3]{3x+2}+x^2}]=0$

Vì $4+\frac{3x^2}{\sqrt[3]{(3x+2)^2}+x\sqrt[3]{3x+2}+x^2}>0$

$=>x^3-3x-2=0<=>...$