Question

[Kem Su]

Giải PT : \(2x^2+x+3=3x\sqrt{x+3}\)

Answer 1

[ღ๖ۣۜLinh's ๖ۣۜLinh'sღ] ★we are one★]

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

\(ĐK:x\ge-3\)

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

Đặt \(\sqrt{x+3}=a,x=b\left(a\ge0\right)\)

\(\left(1\right)\Leftrightarrow a^2+2b^2=3ab\)

\(\Leftrightarrow\left(a-b\right)\left(a-2b\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=b\\a=2b\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x+3}=x\\\sqrt{x+3}=2x\end{cases}\Leftrightarrow}\orbr{\begin{cases}x^2-x-3=0\\4x^2-x-3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1\pm\sqrt{13}}{2}\\x=1,x=\frac{3}{4}\end{cases}}\)

Vậy............

Answer 2

[Trí Tiên]

\(Đkxđ:x\ge-3\)

Với đk trên pt trở thành:  \(2\left(x\right)^2-3\left(x\right)\left(\sqrt{x+3}\right)+\left(\sqrt{x+3}\right)^2=0\)

\(\Leftrightarrow2\left(x\right)^2-2\left(x\right)\left(\sqrt{x+3}\right)-\left(x\right)\left(\sqrt{x+3}\right)+\left(\sqrt{x+3}\right)^2=0\)

\(\Leftrightarrow2x\left(x-\sqrt{x+3}\right)-\sqrt{x+3}\left(x-\sqrt{x+3}\right)=0\)

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

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

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

\(\Leftrightarrow x\ge0\) Hoặc \(\orbr{\begin{cases}x=\frac{1+\sqrt{13}}{2}\\x=\frac{1-\sqrt{13}}{2}\end{cases}}\Leftrightarrow x=\frac{1+\sqrt{13}}{1}\)

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

\(\Leftrightarrow x\ge0\) Hoặc \(\orbr{\begin{cases}x=1\\x=-\frac{3}{4}\end{cases}}\Leftrightarrow x=1\)

So với đk ta đc tập nghiệm: \(S=\left\{1;\frac{1+\sqrt{13}}{2}\right\}\)