đkxđ \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\).
K viết lại đề nữa nhé
\(B=\left(\frac{2x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}-1}\right):\left(1-\frac{x-2}{x+\sqrt{x}+1}\right)\)
\(\Leftrightarrow\frac{2x+1-1\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\frac{x+\sqrt{x}+1-x+2}{x+\sqrt{x}+1}\)
\(\Leftrightarrow\frac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{x+\sqrt{x}+1}{\sqrt{x}+3}\)
\(\Leftrightarrow\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}.\frac{1}{\sqrt{x}+3}\)
\(\Leftrightarrow\frac{\sqrt{x}}{\sqrt{x}+3}\)
\(B.\sqrt{x}=\frac{\sqrt{x}}{\sqrt{x}+3}.\sqrt{x}=\frac{x}{\sqrt{x}+3}\)
\(\frac{x}{\sqrt{x}+3}=\frac{4}{5}\)
\(\Leftrightarrow5x=4\sqrt{x}+12\)
\(\Leftrightarrow5x-4\sqrt{x}-12=0\)
\(\Leftrightarrow5x-10\sqrt{x}+6\sqrt{x}-12=0\)
\(\Leftrightarrow5\sqrt{x}\left(\sqrt{x}-2\right)+6\left(\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\left(5\sqrt{x}-6\right)\left(\sqrt{x}-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}5\sqrt{x}+6=0\\\sqrt{x}+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\sqrt{x}=\frac{-6}{5}\\\sqrt{x}=-2\end{matrix}\right.\) =>k thỏa mãn
=> k tìm dc x thỏa mãn
