Chủ đề này có 1 trả lời

[trang2004]

$\sqrt{2x^3+6}=x+\sqrt{x^2-3x+3}$

[Frosty Flame]

$\sqrt{2x^3+6}=x+\sqrt{x^2-3x+3}$

Xin được sửa đề thành $\sqrt[3]{2x^3+6}=x+\sqrt{x^2-3x+3}$

Pt đã cho

$<=>\sqrt[3]{2x^3+6}-(\frac{2-\sqrt[3]{\frac{9}{2}}}{1+\sqrt[3]{\frac{3}{4}}}x+\frac{\sqrt[3]{\frac{9}{2}}+2\sqrt[3]{\frac{3}{4}}}{1+\sqrt[3]{\frac{3}{4}}})=\sqrt{x^2-3x+3}-(\frac{1-\sqrt[3]{\frac{9}{2}}-\sqrt[3]{\frac{3}{4}}}{1+\sqrt[3]{\frac{3}{4}}}x+\frac{\sqrt[3]{\frac{9}{2}}+2\sqrt[3]{\frac{3}{4}}}{1+\sqrt[3]{\frac{3}{4}}})$

$<=>...<=>(x+\sqrt[3]{\frac{3}{4}})(x-1).A=0($Với $A\neq 0\forall x\in \mathbb{R})$

$<=>x=-\sqrt[3]{\frac{3}{4}};x=1$

Vậy$,...$