a: Ta có: \(\sqrt{\dfrac{2x-3}{x-1}}=2\)
\(\Leftrightarrow\dfrac{2x-3}{x-1}=4\)
\(\Leftrightarrow2x-3=4x-4\)
\(\Leftrightarrow-2x=-1\)
hay \(x=\dfrac{1}{2}\left(nhận\right)\)
b: Ta có: \(\dfrac{\sqrt{2x-3}}{\sqrt{x-1}}=2\)
\(\Leftrightarrow\sqrt{2x-3}=\sqrt{4x-4}\)
\(\Leftrightarrow2x-3=4x-4\)
\(\Leftrightarrow-2x=-1\)
hay \(x=\dfrac{1}{2}\left(loại\right)\)
c: Ta có: \(\sqrt{\dfrac{4x+3}{x+1}}=3\)
\(\Leftrightarrow\dfrac{4x+3}{x+1}=9\)
\(\Leftrightarrow4x+3=9x+9\)
\(\Leftrightarrow-5x=6\)
hay \(x=-\dfrac{6}{5}\left(nhận\right)\)
d: Ta có: \(\dfrac{\sqrt{4x+3}}{\sqrt{x+1}}=3\)
\(\Leftrightarrow4x+3=9x+9\)
\(\Leftrightarrow x=-\dfrac{6}{5}\left(loại\right)\)
a) \(\sqrt{\dfrac{2x-3}{x-1}}=2\left(đk:\dfrac{2x-3}{x-1}\ge0,x\ne1\right)\Leftrightarrow\left(\sqrt{\dfrac{2x-3}{x-1}}\right)^2=4\Leftrightarrow\dfrac{2x-3}{x-1}=4\Leftrightarrow2x-3=4x-4\Leftrightarrow x=\dfrac{1}{2}\)b) \(\dfrac{\sqrt{2x-3}}{\sqrt{x-1}}=2\left(đk:x\ge\dfrac{3}{2},x\ne1\right)\Leftrightarrow\dfrac{2x-3}{x-1}=4\Leftrightarrow2x-3=4x-4\Leftrightarrow x=\dfrac{1}{2}\)(loại)
Vậy \(S=\varnothing\)
