Question

[lý canh hy]

Giải pt

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

Answer 1

[Nguyễn Tất Đạt]

\(x^3+2x^2+2x+3=\left(3x+1\right)\sqrt{x^3+3}\)(ĐKXĐ: \(x\ge-\sqrt[3]{3}\))

Đặt \(\sqrt{x^3+3}=y\)(\(y\ge0\)) \(\Leftrightarrow x^3+3=y^2\). Khi đó pt cho mang dạng:

\(y^2+2x^2+2x=\left(3x+1\right)y\)

\(\Leftrightarrow y^2-3xy+2x^2+2x-y=0\)

\(\Leftrightarrow y^2-xy-2xy+2x^2-\left(y-2x\right)=0\)

\(\Leftrightarrow y\left(y-x\right)-2x\left(y-x\right)-\left(y-2x\right)=0\)

\(\Leftrightarrow\left(y-x\right)\left(y-2x\right)-\left(y-2x\right)=0\)

\(\Leftrightarrow\left(y-2x\right)\left(y-x-1\right)=0\Leftrightarrow\orbr{\begin{cases}y-2x=0\\y-x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}y=2x\\y=x+1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}\sqrt{x^3+3}=2x\left(1\right)\\\sqrt{x^3+3}=x+1\left(2\right)\end{cases}}\)

Ta có: \(\left(1\right)\Leftrightarrow\hept{\begin{cases}x\ge0\\x^3+3=4x^2\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge0\\x^3-x^2-3x^2+3=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\ge0\\x^2\left(x-1\right)-3\left(x-1\right)\left(x+1\right)=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x\ge0\\\left(x-1\right)\left(x^2-3x-3\right)=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\ge0\\\left(x-1\right)\left[\left(x-\frac{3}{2}\right)^2-\frac{21}{4}\right]=0\end{cases}}\)\(\Leftrightarrow x=1\) hoặc \(\orbr{\begin{cases}x=\frac{\sqrt{21}+3}{2}\\x=\frac{3-\sqrt{21}}{2}\end{cases}}\) 

\(\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{\sqrt{21}+3}{2}\end{cases}}\) (loại TH \(x=\frac{3-\sqrt{21}}{2}< 0\))

Lại có: \(\left(2\right)\Leftrightarrow\hept{\begin{cases}x+1\ge0\\x^3+3=x^2+2x+1\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge-1\\x^3-x^2-2x+2=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\ge-1\\x^2\left(x-1\right)-2\left(x-1\right)=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ge-1\\\left(x-1\right)\left(x^2-2\right)=0\end{cases}}\)

\(\Leftrightarrow x=1\) hoặc \(x=\pm\sqrt{2}\)\(\Rightarrow\orbr{\begin{cases}x=1\\x=\sqrt{2}\end{cases}}\)(t/m ĐK) (loại TH \(x=-\sqrt{2}\) vì \(-\sqrt{2}< -1\))

Vậy tập nghiệm của pt là \(S=\left\{1;\sqrt{2};\frac{\sqrt{21}+3}{2}\right\}.\)