Đáp án:
b. x=26
Giải thích các bước giải:
\(\begin{array}{l}
a.{\left[ {x\left( {x - 3} \right)} \right]^2} - 2x\left( {x - 3} \right) - 8 = 0\\
Dat:x\left( {x - 3} \right) = t\\
Pt \to {t^2} - 2t - 8 = 0\\
\to {t^2} + 2t - 4t - 8 = 0\\
\to t\left( {t + 2} \right) - 4\left( {t + 2} \right) = 0\\
\to \left[ \begin{array}{l}
t + 2 = 0\\
t - 4 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
t = - 2\\
t = 4
\end{array} \right.\\
\to \left[ \begin{array}{l}
{x^2} - 3x = - 2\\
{x^2} - 3x = 4
\end{array} \right.\\
\to \left[ \begin{array}{l}
\left( {x - 2} \right)\left( {x - 1} \right) = 0\\
\left( {x - 4} \right)\left( {x + 1} \right) = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\\
x = 1\\
x = 4\\
x = - 1
\end{array} \right.\\
b.DK:x \ge 1\\
x - 2\sqrt {x - 1} = 16\\
\to x - 16 = 2\sqrt {x - 1} \\
\to \left\{ \begin{array}{l}
x \ge 16\\
{x^2} - 32x + 256 = 4x - 4
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \ge 16\\
{x^2} - 26x - 10x + 256 = 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \ge 16\\
\left[ \begin{array}{l}
x - 26 = 0\\
x - 10 = 0
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 10\left( l \right)\\
x = 26\left( {TM} \right)
\end{array} \right.
\end{array}\)
