$5(1+\sqrt{1+x^{3}})= x^{2}.(4x^{2}-25x+18)$
chịu luôn mình ko biết đề thế nào để như của bạn là x^3 ở ngoài căn thì đề quá dễ –  ★·.·´¯`·.·★Poseidon★·.·´¯`·.·★ 26-05-16 03:59 PM
bài thế này à bạn??? –  ★·.·´¯`·.·★Poseidon★·.·´¯`·.·★ 26-05-16 03:56 PM
                        $Nhớ....................Vote....................Nha$
                                               
Dk: $x\ge -1=>x+1\ge 0$
$pt\iff 5(\sqrt{1+x^3}-2x-2)=4x^4-25x^3+18x^2-10x-15$
$\iff \frac{5(1+x^3-4x^2-8x-4)}{\sqrt{1+x^3}+2x+2}=(x^2-5x-3)(4x^2-5x+5)$
$\iff \frac{5(x+1)(x^2-5x-3)}{\sqrt{1+x^3}+2x+2}=(x^2-5x-3)(4x^2-5x+5)$
$x^2-5x-3=0.v.\frac{5(x+1)}{\sqrt{1+x^3}+2x+2}=4x^2-5x+5 $
$Th1: x^2-5x-3=0=>.....$
Ta có: $4x^2-5x+5=(2x-\frac{5}{4})^2+\frac{55}{16}>3$
$Th2:\frac{5(x+1)}{\sqrt{1+x^3}+2x+2}=4x^2-5x+5>3=>5(x+1)>3\sqrt{1+x^3}+6x+6$
$\iff 0>3\sqrt{1+x^3}+x+1(Vo.li)=>Vo.nghiem$
Vậy $x=\frac{5+\sqrt{37}}{2},x=\frac{5-\sqrt{37}}{2}$

ta có 
pt$\Leftrightarrow 4x^{4}-25x^{3}+18x^{2}-5-5\sqrt{1+x^{3}}=0$
   $\Leftrightarrow  4x^{4}-25x^{3}+18x^{2}-10x-15-5(\sqrt{1+x^{3}}-2(x+1))=0$
   $\Leftrightarrow (x^{2}-5x-3)(4x^{2}-5x+5-\frac{5(x+1)}{\sqrt{1+x^{3}}+2(x+1)})=0$
   $\Leftrightarrow  x^{2}-5x-3=0$
hoặc $4x^{2}-5x+5=\frac{5(x+1)}{\sqrt{1+x^{3}}+2(x+1)}$ (1)
VT(1) $\geq \frac{55}{16}>3$
VP= $\frac{5}{2}-\frac{5\sqrt{1+x^{3}}}{2(\sqrt{1+x^{3}}+2(x+1)}\leq \frac{5}{2}$
$\Rightarrow$ (1) VNo
$\Rightarrow x=\frac{5\pm \sqrt{37}}{2}$

 
       

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