\(A=\dfrac{1}{\sqrt{2}}\)
\(\Leftrightarrow\sqrt{\dfrac{2x+3}{x-3}}=\dfrac{1}{\sqrt{2}}\) ( ĐK : \(x\ne3\) )
\(\Leftrightarrow\) \(\dfrac{2x+3}{x-3}=\dfrac{1}{2}\)
\(\Leftrightarrow\) \(\dfrac{2\left(2x+3\right)}{2\left(x-3\right)}=\dfrac{x-3}{2\left(x-3\right)}\)
\(\Leftrightarrow2\left(2x+3\right)=x-3\)
\(\Leftrightarrow4x+6=x-3\)
\(\Leftrightarrow3x=-9\)
\(\Leftrightarrow x=-3\) ( Thỏa mãn )
Vậy \(x=-3\) thì \(A=\dfrac{1}{\sqrt{2}}\)
ĐKXĐ: x khác 3; (2x + 3) và (x - 3) cùng dấu
\(A=\dfrac{1}{\sqrt{2}}\)
\(\Leftrightarrow\sqrt{\dfrac{2x+3}{x-3}}=\dfrac{1}{\sqrt{2}}\)
\(\Leftrightarrow\dfrac{2x+3}{x-3}=\dfrac{1}{2}\) (bình phương hai vế)
\(\Leftrightarrow4x+6=x-3\)
\(\Leftrightarrow3x=-9\)
\(\Leftrightarrow x=-3\)(thỏa mãn ĐKXĐ)