Question

[Lê Đức Lương]

Giải phương trình \(x^3+\left(3x^2-4x-4\right)\sqrt{x+1}=0\)

Answer 1

[Xyz OLM]

ĐKXĐ : x \(\ge-1\)

\(x^3+\left(3x^2-4x-4\right)\sqrt{x+1}=0\)

<=> \(x^3+3x^2\sqrt{x+1}-4\left(x+1\right)\sqrt{x+1}=0\)

<=> \(x^3+3x^2\sqrt{x+1}-4\left(\sqrt{x+1}\right)^3=0\)

<=> \(\left(x^3-x^2\sqrt{x+1}\right)+4\left[x^2\sqrt{x+1}-\left(\sqrt{x+1}\right)^3\right]=0\)

\(\Leftrightarrow x^2\left(x-\sqrt{x+1}\right)+4\sqrt{x+1}\left[x^2-\left(\sqrt{x+1}\right)^2\right]=0\)

<=> \(x^2\left(x-\sqrt{x+1}\right)+4\sqrt{x+1}\left(x-\sqrt{x+1}\right)\left(x+\sqrt{x+1}\right)=0\)

<=> \(\left(x-\sqrt{x+1}\right)\left(x^2+4x\sqrt{x+1}+4x+4\right)=0\)

<=> \(\left(x-\sqrt{x+1}\right)\left(x+2\sqrt{x+1}\right)^2=0\)

<=> \(\left[{}\begin{matrix}x=\sqrt{x+1}\left(1\right)\\x=-2\sqrt{x+1}\left(2\right)\end{matrix}\right.\)

Giải (1) ta có \(x=\sqrt{x+1}\Leftrightarrow\left\{{}\begin{matrix}x^2=x+1\\x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=\dfrac{\sqrt{5}+1}{2}\\x=\dfrac{1-\sqrt{5}}{2}\left(\text{loại}\right)\end{matrix}\right.\\x\ge0\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{\sqrt{5}+1}{2}\)

Giải (2) ta có : \(x=-2\sqrt{x+1}\Leftrightarrow\left\{{}\begin{matrix}x^2-4x-4=0\\x\ge-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\pm\sqrt{8}+2\\x\ge-1\end{matrix}\right.\Leftrightarrow x=\sqrt{8}+2\)

Answer 2

[missing you =]

\(x^3+\left(3x^2-4x-4\right)\sqrt{x+1}=0\left(đk:x\ge-1\right)\)

\(\Leftrightarrow x^3+3x^2\sqrt{x+1}-4\left(x+1\right)\sqrt{x+1}=0\)

\(\Leftrightarrow x^3+3x^2\sqrt{x+1}-4\sqrt{x+1}^3=0\left(1\right)\)

\(TH:x=-1\Rightarrow\left(1\right)\Leftrightarrow-1=0\left(ktm\right)\)

\(TH:x>-1\Rightarrow\left(1\right)\Leftrightarrow\left(\dfrac{x}{\sqrt{x+1}}\right)^3+3\left(\dfrac{x}{\sqrt{x+1}}\right)^2-4=0\)

\(đặt:\dfrac{x}{\sqrt{x+1}}=a\Rightarrow a^3+3a^2-4=0\Leftrightarrow\left(a+2\right)^2\left(a-1\right)=0\Leftrightarrow\left[{}\begin{matrix}a=1=\dfrac{x}{\sqrt{x+1}}\Leftrightarrow\sqrt{x+1}=x\left(2\right)\\a=-2=\dfrac{x}{\sqrt{x+1}}\Leftrightarrow2\sqrt{x+1}=-x\left(3\right)\end{matrix}\right.\)

\(\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2=x+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left[{}\begin{matrix}x=\dfrac{1+\sqrt{5}}{2}\\x=\dfrac{1-\sqrt{5}}{2}\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow x=\dfrac{1+\sqrt{5}}{2}\)

\(\left(3\right)\Leftrightarrow\left\{{}\begin{matrix}-1< x\le0\\4\left(x+1\right)=x^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-1< x\le0\\\left[{}\begin{matrix}x=2+2\sqrt{2}\\x=2-2\sqrt{2}\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow x=2-2\sqrt{2}\)