7 replies to this topic

[lilolilo]

$2\sqrt{x^2-2x-1}+\sqrt[3]{y^3-14}=x-2$

[vuvanquya1nct]

$2\sqrt{x^2-2x-1}+\sqrt[3]{y^3-14}=x-2$

$x$ hay là $y$ vậy ??????

[lilolilo]

dạ x. e nhầm

[lilolilo]

$x$ hay là $y$ vậy ??????

dạ x

[lilolilo]

dạ x á. giải giúp e với

[vuvanquya1nct]

dạ x

Nếu là x thì:

DK:$\begin{bmatrix} x\leq 1-\sqrt{2} & \\ x\geq 1+\sqrt{2} & \end{bmatrix}$

$PT\Leftrightarrow 2\sqrt{x^2-2x-1}+[\sqrt[3]{x^3-14}-(x-2)]=0$

OK

Edited by vuvanquya1nct, 21-01-2014 - 22:14.

[lilolilo]

làm rõ hơn đi ạ

[vuvanquya1nct]

làm rõ hơn đi ạ

$2\sqrt{x^2-2x-1}+[\sqrt[3]{x^3-14}-(x-2)]=0$$

$\Leftrightarrow 2\sqrt{x^2-2x-1}+\frac{(x^3-14)-(x-2)^3}{(\sqrt[3]{x^3-14})^2+(x-2)\sqrt[3]{x^3-14}+(x-2)^3}=0$$

$\Leftrightarrow 2\sqrt{x^2-2x-1}+\frac{6x^2-12x-6}{(\sqrt[3]{x^3-14})^2+(x-2)\sqrt[3]{x^3-14}+(x-2)^3}=0$$

$\sqrt{x^2-2x-1}=0$

Còn $2+\frac{6}{(\sqrt[3]{x^3-14})^2+(x-2)\sqrt[3]{x^3-14}+(x-2)^3}>0$

Edited by vuvanquya1nct, 21-01-2014 - 22:21.