Đáp án:
c) x=10
Giải thích các bước giải:
\(\begin{array}{l}
b)DK:x \ge 4\\
x - 4 = \sqrt {2x - 5} \\
\to {x^2} - 8x + 16 = 2x - 5\\
\to {x^2} - 10x + 21 = 0\\
\to \left[ \begin{array}{l}
x = 7\left( {TM} \right)\\
x = 3\left( l \right)
\end{array} \right.\\
e)DK:x \ge \dfrac{7}{2}\\
\sqrt {{x^2} - 3x - 1} = 2x - 7\\
\to {x^2} - 3x - 1 = 4{x^2} - 28x + 49\\
\to 3{x^2} - 25x + 50 = 0\\
\to \left[ \begin{array}{l}
x = 5\left( {TM} \right)\\
x = \dfrac{{10}}{3}\left( l \right)
\end{array} \right.\\
c)13 - x = \sqrt {x - 1} \\
\to 169 - 26x + {x^2} = x - 1\left( {DK:13 \ge x \ge 1} \right)\\
\to {x^2} - 27x + 170 = 0\\
\to \left[ \begin{array}{l}
x = 17\left( l \right)\\
x = 10\left( {TM} \right)
\end{array} \right.\\
f)\sqrt {{x^4} + 12} = {x^2} + 2\\
\to {x^4} + 12 = {x^4} + 4{x^2} + 4\\
\to 4{x^2} = 8\\
\to {x^2} = 2\\
\to \left[ \begin{array}{l}
x = \sqrt 2 \\
x = - \sqrt 2
\end{array} \right.
\end{array}\)
