(3)

$x^3+x^2-2x-1=(x+1)\sqrt[3]{3x+1}$
lập phương lên cho nó thốn –  ๖ۣۜJinღ๖ۣۜKaido 07-08-16 06:28 PM
$Pt\Leftrightarrow x^3+x^2-3x+x-1=(x+1)\sqrt[3]{3x+1}$

Đặt $\sqrt[3]{3x+1}=t \Rightarrow 3x=t^3-1$.  Phương trình trở thành: 

$x^3+x^2-(t^3-1)+x-1=(x+1)t\Leftrightarrow  x^3-t^3+x^2-xt+x-t=0\Leftrightarrow (x-t)(x^2+xt+t^2+x+1)=0$
 
Dễ chứng minh  $x^2+xt+t^2+x+1>0$
 Do đó: $x=t\Leftrightarrow x=\sqrt[3]{3x+1}\Leftrightarrow ...............$
dạng chuẩn để dùng Cacnado còn cái j nữa! –  ★·.·´¯`·.·★Poseidon★·.·´¯`·.·★ 07-08-16 08:59 PM
dùng công thức Cacnado đi –  ★·.·´¯`·.·★Poseidon★·.·´¯`·.·★ 07-08-16 08:58 PM
quan trọng là phần sau giải ntn :)) –  tran85295 07-08-16 08:53 PM

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