a) \(3+\sqrt{2x-3}=x\) (ĐK: \(x\ge3\))
\(\Leftrightarrow\sqrt{2x-3}=x-3\)
\(\Leftrightarrow2x-3=\left(x-3\right)^2\)
\(\Leftrightarrow2x-3=x^2-6x+9\)
\(\Leftrightarrow x^2-8x+12=0\)
\(\Leftrightarrow\left(x-6\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\left(tm\right)\\x=2\left(ktm\right)\end{matrix}\right.\)
b) \(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{3-11\sqrt{x}}{9-x}=\dfrac{6}{\sqrt{x}-3}\) \(\left(x\ge0;x\ne9\right)\)
\(\Leftrightarrow\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{6\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow2x-6\sqrt{x}+x+3\sqrt{x}+\sqrt{x}+3+11\sqrt{x}-3=6\sqrt{x}+18\)
\(\Leftrightarrow3x+9\sqrt{x}=6\sqrt{x}+18\)
\(\Leftrightarrow3x+3\sqrt{x}-18=0\)
\(\Leftrightarrow3\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)=0\)
\(\Leftrightarrow\sqrt{x}=2\)
\(\Leftrightarrow x=4\left(tm\right)\)
c) \(x^2+4x+7=\left(x+4\right)\sqrt{x^2+7}\)
\(\Leftrightarrow\dfrac{x^2+4x+7}{x+4}=\sqrt{x^2+7}\)
\(\Leftrightarrow\dfrac{x^2+4x+7}{x+4}-4=\sqrt{x^2+7}-4\)
\(\Leftrightarrow\dfrac{x^2-9}{x+4}-\dfrac{x^2-9}{\sqrt{x^2+7}+4}=0\)
\(\Leftrightarrow\left(x^2-9\right)\left(\dfrac{1}{x+4}-\dfrac{1}{\sqrt[]{x^2+7}+4}\right)=0\)
\(\Leftrightarrow x^2=9\)
\(\Leftrightarrow x=\pm3\)
