Ở phân thức thứ 2 không phải \(\sqrt{x-2}\) đâu mà là \(\sqrt{x}-2\)

Từ điều kiện \(x-3\sqrt{x}+2=0\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}=2\end{matrix}\right.\). Nhưng ta thấy rõ \(\sqrt{x}=2\) là điều không thể do điều kiện xác định của K. Ta chỉ nhận giá trị \(\sqrt{x}=1\Leftrightarrow x=1\)

Mặt khác \(K=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)

\(K=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(K=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\) 

\(K=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(K=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

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

Mà \(\sqrt{x}=1\) nên \(K=\dfrac{1+1}{1-3}=-1\)

Vậy tại \(x-3\sqrt{x}+2=0\) thì \(K=-1\)

đk x>=0 ; x khác 4 ; 9 

\(K=\dfrac{2\sqrt{x}-9-\left(x-9\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

1, \(=\dfrac{\sqrt{6}\left(\sqrt{6}-1\right)}{\sqrt{6}-1}+\dfrac{\sqrt{6}\left(\sqrt{6}+1\right)}{\sqrt{6}}=\sqrt{6}+\sqrt{6}+1=2\sqrt{6}+1\)

2, \(=\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}}+\dfrac{\sqrt{3}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}=\sqrt{3}+1-\sqrt{3}=1\)

ĐKXĐ : \(x\ge-4\)

(2x + 1)² = (x + 4)\(\sqrt{4x^2+1}\)

<=> \(4x^2+1+4x=x\sqrt{4x^2+1}+4\sqrt{4x^2+1}\)

<=> \(\left(\sqrt{4x^2+1}-4\right)\left(\sqrt{4x^2+1}-x\right)=0\)

<=> \(\left[{}\begin{matrix}\sqrt{4x^2+1}=4\\\sqrt{4x^2+1}=x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x^2=15\\\left\{{}\begin{matrix}3x^2+1=0\\x\ge0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\dfrac{\sqrt{15}}{2}\left(tm\right)\\∄x\end{matrix}\right.\Leftrightarrow x=\pm\dfrac{\sqrt{15}}{2}\)

\(\sqrt{\dfrac{3x^2-3x-2x+2}{4x^2-12x+9}}=\sqrt{\dfrac{\left(3x-2\right)\left(x-1\right)}{\left(2x-3\right)^2}}\)

đk \(\left\{{}\begin{matrix}\left(3x-2\right)\left(x-1\right)>0\\2x-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>1\\x< \dfrac{2}{3}\\x\ne\dfrac{3}{2}\end{matrix}\right.\)

\(=\sqrt{2}-1-\sqrt{2}-1-2=-4\)

\(ĐKXĐ:x\ge0\)

\(4x-2\sqrt{x}=0\)

\(\Leftrightarrow2\sqrt{x}\left(2\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\\2\sqrt{x}-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=\dfrac{1}{4}\left(tm\right)\end{matrix}\right.\)

đk x >= 0 

\(2\sqrt{x}\left(2\sqrt{x}-1\right)=0\Leftrightarrow x=0;x=\dfrac{1}{4}\left(tm\right)\)