ĐKXĐ: ...
\(\Leftrightarrow x^3-3x\left(x+1\right)+2\sqrt{\left(x+1\right)^3}=0\)
Đặt \(\left\{{}\begin{matrix}x=a\\\sqrt{x+1}=b\end{matrix}\right.\)
\(\Rightarrow a^3-3ab^2+2b^3=0\)
\(\Leftrightarrow\left(a+2b\right)\left(a-b\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}2b=-a\\a=b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x+1}=-x\left(x\le0\right)\\x=\sqrt{x+1}\left(x\ge0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x-4=0\\x^2-x-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2-2\sqrt{2}\\x=\frac{1+\sqrt{5}}{2}\end{matrix}\right.\)
Lời giải:
ĐKXĐ: $x\geq -1$
Đặt $\sqrt{x+1}=a(a\geq 0)$ thì PT trở thành:
$x^3-3x(x+1)+2\sqrt{(x+1)^3}=0$
$\Leftrightarrow x^3-3xa^2+2a^3=0$
$\Leftrightarrow (x^3-xa^2)-(2xa^2-2a^3)=0$
$\Leftrightarrow x(x-a)(x+a)-2a^2(x-a)=0$
$\Leftrightarrow (x-a)(x^2+ax-2a^2)=0$
$\Leftrightarrow (x-a)[(x+a)(x-a)+a(x-a)]=0$
$\Leftrightarrow (x-a)^2(x+2a)=0$
Nếu $x-a=0$
$\Rightarrow x^2=a^2\Leftrightarrow x^2=x+1$
$\Rightarrow x=\frac{1\pm \sqrt{5}}{2}$. Vì $x=a\geq 0$ nên $x=\frac{1+\sqrt{5}}{2}$
Nếu $x+2a=0$
$\Rightarrow x^2=4a^2\Leftrightarrow x^2=4(x+1)$
$\Rightarrow x=2\pm 2\sqrt{2}$. Mà $x=-2a\leq 0$ nên $x=2-2\sqrt{2}$
Vậy..........
