4 replies to this topic

[dattoan]

$\frac{x^{2}}{2}+\frac{x}{2}+1=\sqrt{2x^{3}-x^{2}+x+1}$

Edited by dattoan, 03-08-2016 - 00:36.

[loolo]

Đk: $x\geq \frac{-1}{2}$

pt đã cho tương đương với 

$x^{2}+x+2=2\sqrt{2x^{3}-x^{2}+x+1}$

$\Leftrightarrow (x^{2}+x+2)^{2}=4.(2x^{3}-x^{2}+x+1)$

$\Leftrightarrow$ $x^{4}-6x^{3}+9x^{2}=0$

$\Leftrightarrow x^{2}(x^{2}-6x+9)=0$

$\Leftrightarrow x^{2}(x-3)^{2}=0$

[nganha2001]

$\frac{x^{2}}{2}+\frac{x}{2}+1=\sqrt{2x^{3}-x^{2}+x+1}$

đk:   $x\geq \frac{-1}{2}$

$\Leftrightarrow x^{2}+x+2=2\sqrt{2x^{3}-x^{2}+x+1}$

$\Leftrightarrow (x^{2}-x+1)+(2x+1)=2\sqrt{(2x+1)(x^{2}-x+1)}$

$\Leftrightarrow (\sqrt{x^{2}-x+1}-\sqrt{2x+1})^{2}=0$

$\Leftrightarrow \sqrt{x^{2}-x+1}=\sqrt{2x+1}$

[Zeref]

$\frac{x^{2}}{2}+\frac{x}{2}+1=\sqrt{2x^{3}-x^{2}+x+1}$

PT ban đầu

$<=>x^2+x+2-2\sqrt{2x^{3}-x^{2}+x+1}=0$

$<=>(\sqrt{2x^{3}-x^{2}+x+1}-2x-1)(\frac{\sqrt{2x^{3}-x^{2}+x+1}}{2x+1}-1)=0$

$<=>...$

[dattoan]

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