Đáp án:
x=1
Giải thích các bước giải:
\(\begin{array}{l}
x - 1 = \left( {\sqrt x + 3} \right).\dfrac{{\sqrt x - 5}}{{\sqrt x + 3}} + 2\sqrt {x + 3} \\
\to x - 1 = \sqrt x - 5 + 2\sqrt {x + 3} \\
\to x - \sqrt x + 4 = 2\sqrt {x + 3} \\
\to {\left( {x - \sqrt x + 4} \right)^2} = 4\left( {x + 3} \right)\\
\to {x^2} + x + 16 - 2x\sqrt x + 8x - 8\sqrt x = 4x + 12\\
\to {x^2} - 2x\sqrt x + 5x - 8\sqrt x + 4 = 0\\
\to {x^2} - x\sqrt x - x\sqrt x + x + 4x - 4\sqrt x - 4\sqrt x + 4 = 0\\
\to x\sqrt x \left( {\sqrt x - 1} \right) - x\left( {\sqrt x - 1} \right) + 4\sqrt x \left( {\sqrt x - 1} \right) - 4\left( {\sqrt x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
\sqrt x - 1 = 0\\
x\sqrt x - x + 4\sqrt x - 4 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
x\left( {\sqrt x - 1} \right) + 4\left( {\sqrt x - 1} \right) = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
\left( {\sqrt x - 1} \right)\left( {x + 4} \right) = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
x = - 4\left( l \right)
\end{array} \right.\\
KL:x = 1
\end{array}\)
