\(\dfrac{\sqrt{x}-2}{\sqrt{x}-1}=\dfrac{\sqrt{x}}{\sqrt{x}+1}\) (ĐK: \(x\ge0,x\ne1\))
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)=\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)\)
\(\Leftrightarrow x-\sqrt{x}=x-2\sqrt{x}+\sqrt{x}-2\)
\(\Leftrightarrow x-\sqrt{x}=x-\sqrt{x}-2\)
\(\Leftrightarrow x-x=\sqrt{x}-\sqrt{x}-2\)
\(\Leftrightarrow0=-2\) (vô lý)
⇒ Phương trình vô nghiệm
\(\dfrac{\sqrt{x}-2}{\sqrt{x}-1}=\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
\(\Rightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)=\sqrt{x}\left(\sqrt{x}-1\right)\)
\(\Leftrightarrow x+\sqrt{x}-2\sqrt{x}-2=x-\sqrt{x}\)
\(\Leftrightarrow-2=0\) (sai)
\(đk:x\ge0;x\ne1\)
\(\dfrac{\sqrt{x}-2}{\sqrt{x}-1}=\dfrac{\sqrt{x}}{\sqrt{x}+1}\\ \Rightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)=\sqrt{x}\left(\sqrt{x}-1\right)\\ \Rightarrow x-2\sqrt{x}+\sqrt{x}-2=x-\sqrt{x}\\ \Rightarrow-\sqrt{x}-2+\sqrt{x}=0\\ \Rightarrow-2=0\left(voli\right)\)
Vậy phương trình vô nghiệm
