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$x^{2}-3x+3=(4+3x-\frac{4}{x})\sqrt{x-1}$

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#1
Diepnguyencva

Diepnguyencva

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1) $x^{2}-3x+3=(4+3x-\frac{4}{x})\sqrt{x-1}$

2) $x^{2}\sqrt[4]{2-x^{^{4}}}-1=x^{4}-x^{3}$



#2
Le Tuan Canhh

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1) $PT \Leftrightarrow x^{3}-3x^{2}+3x=(3x^{2}+4x-4)\sqrt{x-1}\Leftrightarrow x^{3}-3x(x-1)-3x^{2}\sqrt{x-1}-4(x-1)\sqrt{x-1}=0$

Đặt $t=\frac{x}{\sqrt{x-1}}$

PT mới : $t^{3}-3t^{2}-3t-4=0\Leftrightarrow t=4\Rightarrow \frac{x}{\sqrt{x-1}}=4\Leftrightarrow x=2$

 

P/s: đáp lễ với khóa trên  :D


Edited by Le Tuan Canhh, 18-10-2022 - 15:59.

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