ĐKXĐ : \(x\ge2\)
Với \(A=\dfrac{x+3}{\sqrt{x}}\)
Khi đó \(A\sqrt{x}+x-1=2\sqrt{3x}+2\sqrt{x-2}\)
<=> \(\dfrac{x+3}{\sqrt{x}}.\sqrt{x}+x-1=2\sqrt{3x}+2\sqrt{x-2}\)
<=> \(x+1=\sqrt{3x}+\sqrt{x-2}\)
Đặt \(\sqrt{3x}=a;\sqrt{x-2}=b\left(a>0;b\ge0\right)\)
Khi đó \(a^2-b^2=2\left(x+1\right)\Leftrightarrow\dfrac{a^2-b^2}{2}=x+1\)
PT trở thành \(\dfrac{a^2-b^2}{2}=a+b\)
<=> \(\left(a+b\right)\left(\dfrac{a-b}{2}-1\right)=0\)
<=> \(\dfrac{a-b}{2}-1=0\left(a+b>0\right)\)
<=> a = b + 2
Khi đó \(\sqrt{3x}=\sqrt{x-2}+2\)
<=> \(\left\{{}\begin{matrix}3x=x+2+4\sqrt{x-2}\\x\ge2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=2\sqrt{x-2}\\x\ge2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-2x+1=4\left(x-2\right)\\x\ge2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-3\right)^2=0\\x\ge2\end{matrix}\right.\Leftrightarrow x=3\)(tm)
\(\)
