Question

giải pt

\(\sqrt{\dfrac{2x-3}{x-1}}=2\)

Answer 1

\(\sqrt{\dfrac{2x-3}{x-1}}=2\) ( 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...

Answer 2

\(\sqrt{\dfrac{2x-3}{x-1}}=2\)

\(\Leftrightarrow\dfrac{2x-3}{x-1}=4\)

\(\Leftrightarrow2x-3=4\left(x-1\right)\)

\(\Leftrightarrow2x-3-4\left(x-1\right)=0\)

\(\Leftrightarrow2x-3-4x+4=0\)

\(\Leftrightarrow-2x+1=0\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy \(x=\dfrac{1}{2}\)

Answer 3

ĐKXĐ: \(x\ne1\)

=>\(\dfrac{2x-3}{x-1}=4\)

<=>\(2x-3=4x-4\)

<=>\(2x-1=0\)

<=>\(x=\dfrac{1}{2}\)(nhận)

Answer 4

ĐKXĐ:x-1 \(\ne0\Leftrightarrow x\ne1\)

Answer 5

=>\(\dfrac{2x-3}{x-1}=4\Leftrightarrow2x-3=4x-4\Leftrightarrow-3+4=4x-2x\Leftrightarrow x=\dfrac{1}{2}\)