19: Ta có: \(\dfrac{3x-5}{\sqrt{x+4}}=\sqrt{x+4}\)
\(\Leftrightarrow3x-5=x+4\)
hay \(x=\dfrac{9}{2}\left(nhận\right)\)
\(14,\sqrt{x+4\sqrt{x-4}}=2\left(x\ge4\right)\\ \Leftrightarrow\sqrt{\left(\sqrt{x-4}+2\right)^2}=2\\ \Leftrightarrow\sqrt{x-4}+2=2\\ \Leftrightarrow\sqrt{x-4}=0\Leftrightarrow x-4=0\Leftrightarrow x=4\left(tm\right)\)
\(15,\sqrt{5x-4}=x-2\left(x\ge\dfrac{4}{5}\right)\\ \Leftrightarrow5x-4=x^2-4x+4\\ \Leftrightarrow x^2-9x+8=0\\ \Leftrightarrow\left(x-8\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=8\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\)
\(16,\sqrt{x^2-2x+4}=2x-2\left(x\in R\right)\\ \Leftrightarrow x^2-2x+4=4x^2-8x+4\\ \Leftrightarrow3x^2-6x=0\Leftrightarrow3x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
\(17,\sqrt{2x^2-2x+1}=2x-1\left(x\in R\right)\\ \Leftrightarrow2x^2-2x+1=4x^2-4x+1\\ \Leftrightarrow2x^2-2x=0\Leftrightarrow2x\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
\(18,\sqrt{14+\sqrt{10x}}=3+\sqrt{5}\left(x\ge0\right)\\ \Leftrightarrow14+\sqrt{10x}=14+6\sqrt{5}\\ \Leftrightarrow\sqrt{10x}=6\sqrt{5}\Leftrightarrow10x=180\Leftrightarrow x=18\left(tm\right)\)
\(19,\dfrac{3x-5}{\sqrt{x+4}}=\sqrt{x+4}\left(x>-4\right)\\ \Leftrightarrow3x-5=x+4\\ \Leftrightarrow2x=9\Leftrightarrow x=\dfrac{9}{2}\left(tm\right)\)
\(20,\sqrt{x-2}=-\sqrt{2}\cdot\sqrt{2}\left(x\ge2\right)\\ \Leftrightarrow\sqrt{x-2}=-\sqrt{4}\\ \Leftrightarrow x-2=4\\ \Leftrightarrow x=6\left(tm\right)\)
