Question

[Phạm Như Nam]

\(x=2\sqrt{x}\)

\(x^2=3\)

\(x^2=15\)

tìm x
 

Answer 1

[Mei Shine]

*ĐKXĐ: \(x\ge0\)

Ta có: \(x=2\sqrt{x}\Leftrightarrow x-2\sqrt{x}=0\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\sqrt{x}=2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\) (tm)

*Ta có: \(x^2=3\Leftrightarrow x^2-3=0\Leftrightarrow\left(x-\sqrt{3}\right)\left(x+\sqrt{3}=0\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\sqrt{3}=0\\x+\sqrt{3}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\)

*Ta có: \(x^2=15\Leftrightarrow x^2-15=0\Leftrightarrow\left(x-\sqrt{15}\right)\left(x+\sqrt{15}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\sqrt{15}=0\\x+\sqrt{15}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{15}\\x=-\sqrt{15}\end{matrix}\right.\)