Question

Tìm x, biết :

a) \(\sqrt{x^2-9}-3\sqrt{x-3}=0\)

b) \(\sqrt{x^2-4}-2\sqrt{x+2}=0\)

Answer 1

Answer 2

a) \(\sqrt{x^2-9}-3\sqrt{x-3}=0\)(ĐKXĐ: \(x\ge\pm3\)

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

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

\(\left[{}\begin{matrix}\sqrt{x-3}=0\\\sqrt{x+3}-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=6\end{matrix}\right.\)(TM)

b)\(\sqrt{x^2-4}-2\sqrt{x+2}=0\)

ĐKXĐ: \(x\ge\pm2\)

\(\Leftrightarrow\sqrt{x-2}.\sqrt{x+2}-2\sqrt{x+2}=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+2}=0\\\sqrt{x-2}-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\left(ktm\right)\\x=6\left(tm\right)\end{matrix}\right.\)