\(\sqrt{\dfrac{2x-3}{x-1}}=2\) ( đk: \(x\ne1\))
\(\Leftrightarrow\dfrac{2x-3}{x-1}=4\)
\(\Leftrightarrow4\left(x-1\right)=2x-3\)
\(\Leftrightarrow4x-4=2x-3\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\dfrac{1}{2}\) (tm)
Vậy...
\(\sqrt{\dfrac{2x-3}{x-1}}=2\)
\(\Leftrightarrow\dfrac{2x-3}{x-1}=4\)
\(\Leftrightarrow2x-3=4\left(x-1\right)\)
\(\Leftrightarrow2x-3-4\left(x-1\right)=0\)
\(\Leftrightarrow2x-3-4x+4=0\)
\(\Leftrightarrow-2x+1=0\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy \(x=\dfrac{1}{2}\)
ĐKXĐ: \(x\ne1\)
=>\(\dfrac{2x-3}{x-1}=4\)
<=>\(2x-3=4x-4\)
<=>\(2x-1=0\)
<=>\(x=\dfrac{1}{2}\)(nhận)
=>\(\dfrac{2x-3}{x-1}=4\Leftrightarrow2x-3=4x-4\Leftrightarrow-3+4=4x-2x\Leftrightarrow x=\dfrac{1}{2}\)
