Đáp án: $x=\dfrac{1-\sqrt{5}}{2}$
Giải thích các bước giải:
ĐKXĐ: $x\ge -1$
Ta có: $a^3+b^3+c^3=(a+b+c)^3-3(a+b)(b+c)(c+a)$ (Hằng đẳng thức)
Với $a=x, b=\sqrt{x+1}, c=\sqrt{2}$
$\to x^3+(\sqrt{x+1})^3+(\sqrt{2})^3=(x+\sqrt{x+1}+\sqrt{2})^3-3(x+\sqrt{x+1})(\sqrt{x+1}+\sqrt{2})(\sqrt{2}+x)$
$\to x^3+(\sqrt{x+1})^3+(\sqrt{2})^3-(x+\sqrt{x+1}+\sqrt{2})^3=-3(x+\sqrt{x+1})(\sqrt{x+1}+\sqrt{2})(\sqrt{2}+x)$
$\to 0=-3(x+\sqrt{x+1})(\sqrt{x+1}+\sqrt{2})(\sqrt{2}+x)$
$\to (x+\sqrt{x+1})(\sqrt{x+1}+\sqrt{2})(\sqrt{2}+x)=0$
$\to x+\sqrt{x+1}=0$
Vì $\sqrt{x+1}+\sqrt{2}>0, \sqrt{2}+x\ge \sqrt{2}-1>0$
$\to (x+1)+\sqrt{x+1}-1=0$
$\to (\sqrt{x+1})^2+\sqrt{x+1}-1=0$
$\to \sqrt{x+1}=\dfrac{-1+\sqrt{5}}{2}$
$\to x+1=(\dfrac{-1+\sqrt{5}}{2})^2$
$\to x=(\dfrac{-1+\sqrt{5}}{2})^2-1$
$\to x=\dfrac{1-\sqrt{5}}{2}$
