\(\sqrt{x-2}+1=2x-\dfrac{20}{x+2}\left(1\right)\)
Đk: \(x\ge2\)
\(\left(1\right)\Leftrightarrow\sqrt{x-2}-1=2x-\dfrac{20}{x+2}-2\)
\(\Leftrightarrow\dfrac{\left(x-2\right)-1}{\sqrt{x-2}+1}=\dfrac{2x\left(x+2\right)-2\left(x+2\right)-20}{x+2}\)
\(\Leftrightarrow\dfrac{\left(x-2\right)-1}{\sqrt{x-2}+1}=\dfrac{2x^2+2x-24}{x+2}\)
\(\Leftrightarrow\dfrac{x-3}{\sqrt{x-2}+1}=\dfrac{2\left(x-3\right)\left(x+4\right)}{x+2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(nhận\right)\\\dfrac{1}{\sqrt{x-2}+1}=2.\dfrac{x+4}{x+2}\left(2\right)\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow2\left(x+4\right)\sqrt{x-2}+2\left(x+4\right)=x+2\)
\(\Leftrightarrow2\left(x+4\right)\sqrt{x-2}+x+6=0\left(3\right)\)
Ta có \(x\ge2>0\Rightarrow2\left(x+4\right)\sqrt{x-2}+x+6>0\)
Vì vậy phương trình (3) vô nghiệm. Khi đó phương trình (2) cũng vô nghiệm.
Vậy phương trình (1) có nghiệm duy nhất là \(x=3\)
