Ta có: \(\dfrac{2\sqrt{x}+3-x}{x-1}=1\)
\(\Leftrightarrow-x+2\sqrt{x}+3=x-1\)
\(\Leftrightarrow-x+2\sqrt{x}+3-x+1=0\)
\(\Leftrightarrow-2x+2\sqrt{x}+4=0\)
\(\Leftrightarrow-2\left(x-\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)=0\)
hay x=4
Đề là ri phải ko bn:\(\dfrac{2\sqrt{x+3}-x}{x-1}=1\)