Đáp án:
d) x=4
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ge 3\\
Pt \to 2x + 1 = 4 + 4\sqrt {x - 3} + x - 3\\
\to x + 4 = 4 + 4\sqrt {x - 3} \\
\to x = 4\sqrt {x - 3} \\
\to {x^2} = 16\left( {x - 3} \right)\\
\to \left[ \begin{array}{l}
x = 12\\
x = 4
\end{array} \right.\left( {TM} \right)\\
c)DK:\left\{ \begin{array}{l}
x \ge 3\\
4 \ge \sqrt {x - 3}
\end{array} \right. \to \left\{ \begin{array}{l}
x \ge 3\\
16 \ge x - 3
\end{array} \right. \to \left\{ \begin{array}{l}
x \ge 3\\
19 \ge x
\end{array} \right.\\
Pt \to 2x + 1 = 16 - 8\sqrt {x - 3} + x - 3\\
\to x + 4 = 16 - 8\sqrt {x - 3} \\
\to 12 - x = 8\sqrt {x - 3} \\
\to 144 - 24x + {x^2} = 64\left( {x - 3} \right)\left( {DK:12 \ge x} \right)\\
\to {x^2} - 88x + 336 = 0\\
\to \left[ \begin{array}{l}
x = 84\left( l \right)\\
x = 4\left( {TM} \right)
\end{array} \right.\\
b)DK:x \ge 3\\
Pt \to \sqrt {3x + 4} = 3 + \sqrt {x - 3} \\
\to 3x + 4 = 9 + 6\sqrt {x - 3} + x - 3\\
\to 2x - 2 = 6\sqrt {x - 3} \\
\to x - 1 = 3\sqrt {x - 3} \\
\to {x^2} - 2x + 1 = 9x - 27\\
\to {x^2} - 11x + 28 = 0\\
\to \left[ \begin{array}{l}
x = 7\\
x = 4
\end{array} \right.\left( {TM} \right)\\
d)DK:5 \ge x \ge 0\\
Pt \to x - 2\sqrt x + 1 = 5 - x\\
\to 2x - 2\sqrt x - 4 = 0\\
\to \left[ \begin{array}{l}
\sqrt x = 2\\
\sqrt x = - 1\left( l \right)
\end{array} \right.\\
\to x = 4\left( {TM} \right)
\end{array}\)
