Đáp án:
e) x=8
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ge 1\\
\sqrt {x + 1} = 1 + \sqrt {x - 1} \\
\to x + 1 = 1 + 2\sqrt {x - 1} + x - 1\\
\to 2\sqrt {x - 1} = 1\\
\to \sqrt {x - 1} = \dfrac{1}{2}\\
\to x - 1 = \dfrac{1}{4}\\
\to x = \dfrac{5}{4}\left( {TM} \right)\\
c)DK:\left[ \begin{array}{l}
x \ge \sqrt 7 \\
x \le - \sqrt 7
\end{array} \right.\\
\sqrt {{x^2} + 9} = 2 + \sqrt {{x^2} - 7} \\
\to {x^2} + 9 = 4 + 4\sqrt {{x^2} - 7} + {x^2} - 7\\
\to 4\sqrt {{x^2} - 7} = 12\\
\to \sqrt {{x^2} - 7} = 3\\
\to {x^2} - 7 = 9\\
\to {x^2} = 16\\
\to \left[ \begin{array}{l}
x = 4\\
x = - 4
\end{array} \right.\left( {TM} \right)\\
e)DK:x \ge - \dfrac{1}{3}\\
\sqrt {x + 1} = 8 - \sqrt {3x + 1} \\
\to \sqrt {x + 1} + \sqrt {3x + 1} = 8\\
\to x + 1 + 2\sqrt {3{x^2} + 4x + 1} + 3x + 1 = 64\\
\to 4x + 2 + 2\sqrt {3{x^2} + 4x + 1} = 64\\
\to 2\sqrt {3{x^2} + 4x + 1} = 62 - 4x\\
\to \sqrt {3{x^2} + 4x + 1} = 31 - 2x\\
\to 3{x^2} + 4x + 1 = 961 - 124x + 4{x^2}\left( {DK:\dfrac{{31}}{2} \ge x} \right)\\
\to {x^2} - 128x + 960 = 0\\
\to \left[ \begin{array}{l}
x = 120\left( l \right)\\
x = 8\left( {TM} \right)
\end{array} \right.
\end{array}\)
