Đáp án: $ x=5\pm\sqrt{33}$
Giải thích các bước giải:
ĐKXĐ: $x>-1$
$10\sqrt{x^3+1}=3(x^2+2)$
$\to 10\sqrt{x^3+1}-(30x+30)=3(x^2-10x-8)$
$\to 10\sqrt{x^3+1}-10(3x+3)=3(x^2-10x-8)$
$\to 10(\sqrt{x^3+1}-(3x+3))=3(x^2-10x-8)$
$\to 10\cdot\dfrac{x^3+1-(3x+3)^2}{\sqrt{x^3+1}+(3x+3)}=3(x^2-10x-8)$
$\to 10\cdot\dfrac{x^3-9x^2-18x-8}{\sqrt{x^3+1}+(3x+3)}=3(x^2-10x-8)$
$\to 10\cdot\dfrac{\left(x+1\right)\left(x^2-10x-8\right)}{\sqrt{x^3+1}+(3x+3)}-3(x^2-10x-8)=0$
$\to x^2-10x-8=0\to x=5\pm\sqrt{33}$
Hoặc $10\cdot\dfrac{x+1}{\sqrt{x^3+1}+(3x+3)}-3=0$
$\to 10\cdot\dfrac{x+1}{\sqrt{(x+1)(x^2-x+1)}+3(x+1)}-3=0$
$\to 10\cdot\dfrac{1}{\sqrt{\dfrac{x^2-x+1}{x+1}}+3}-3=0$
Mà $P=\dfrac{x^2-x+1}{x+1}-(\:2\sqrt{3}-3)$
$\to P=\dfrac{x^2-x+1-(\:2\sqrt{3}-3)(x+1)}{x+1}$
$\to P=\dfrac{\left(x+1-\sqrt{3}\right)^2}{x+1}\ge 0\quad\forall x$
$\to P\ge \:2\sqrt{3}-3$
$\to 10\cdot \dfrac{1}{\sqrt{\dfrac{x^2-x+1}{x+1}}+3}-3\le 10\cdot \dfrac{1}{\sqrt{ \:2\sqrt{3}-3}+3}-3<0$
$\to 10\cdot\dfrac{x+1}{\sqrt{x^3+1}+(3x+3)}-3=0$ vô nghiệm
