Đáp án:7.x=5

8.x=3

Giải thích các bước giải:

\(\begin{array}{l}
7)\sqrt {x + 4}  + \sqrt {x - 4}  = 2x - 12 + 2\sqrt {{x^2} - 16} (*)\\
Dk:x \ge 4\\
Dat:\;t = \sqrt {x + 4}  + \sqrt {x - 4} (t \ge 2\sqrt 2 )\\
 \Rightarrow {t^2} = 2x + 2\sqrt {(x + 4)(x - 4)}  \Rightarrow \sqrt {{x^2} - 16}  = \frac{{{t^2} - 2x}}{2}\\
(*) \Rightarrow t = 2x - 12 + {t^2} - 2x\\
 \Leftrightarrow {t^2} - t - 12 = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
t = 4\\
t =  - 3 \to loai
\end{array} \right.\\
t = 4 \Rightarrow \sqrt {x + 4}  + \sqrt {x - 4}  = 4\\
Dat:\sqrt {x + 4}  = a;\sqrt {x - 4}  = b(a,b \ge 0)\\
 \Rightarrow \left\{ \begin{array}{l}
{a^2} - {b^2} = 8\\
a + b = 4
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
a - b = 2\\
a + b = 4
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
a = 3\\
b = 1
\end{array} \right. \Rightarrow x = 5 \to TM\\
8)\sqrt {2x + 3}  + \sqrt {x + 1}  = 3x + 2\sqrt {(2x + 3)(x + 1)}  - 16(**)\\
Dk:x \ge  - 1\\
Dat:t = \sqrt {2x + 3}  + \sqrt {x + 1} (t \ge 1)\\
 \Rightarrow {t^2} = 3x + 4 + 2\sqrt {(2x + 3)(x + 1)}  \Rightarrow \sqrt {(2x + 3)(x + 1)}  = \frac{{{t^2} - 3x - 4}}{2}\\
(**) \Rightarrow t = 3x + {t^2} - 3x - 4 - 16\\
 \Leftrightarrow {t^2} - t - 20 = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
t =  - 4 \to loai\\
t = 5
\end{array} \right.\\
t = 5 \Rightarrow \sqrt {2x + 3}  + \sqrt {x + 1}  = 5\\
Dat:\left\{ \begin{array}{l}
\sqrt {2x + 3}  = a\\
\sqrt {x + 1}  = b
\end{array} \right.(a,b \ge 0)\\
 \Rightarrow \left\{ \begin{array}{l}
{a^2} - 2{b^2} = 1\\
a + b = 5
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
a = 17\\
b =  - 12
\end{array} \right. \to loai\\
\left\{ \begin{array}{l}
a = 3\\
b = 2
\end{array} \right.
\end{array} \right.\\
 \Rightarrow x = 3 \to TM
\end{array}\)