Đáp án:
3) \(\left[ \begin{array}{l}
x = 4\\
x = - 1
\end{array} \right.\)
4) x=1
Giải thích các bước giải:
\(\begin{array}{l}
3){x^2} + \sqrt {{x^2} - 3x + 5} = 3x + 7\\
\to {x^2} - 3x - 7 + \sqrt {{x^2} - 3x + 5} = 0\\
\to {x^2} - 3x + 5 + \sqrt {{x^2} - 3x + 5} - 12 = 0\\
Đặt:\sqrt {{x^2} - 3x + 5} = t\left( {t \ge 0} \right)\\
Pt \to {t^2} + t - 12 = 0\\
\to \left( {t + 4} \right)\left( {t - 3} \right) = 0\\
\to \left[ \begin{array}{l}
t = - 4\left( l \right)\\
t = 3
\end{array} \right.\\
\to \sqrt {{x^2} - 3x + 5} = 3\\
\to {x^2} - 3x + 5 = 9\\
\to {x^2} - 3x - 4 = 0\\
\to \left( {x - 4} \right)\left( {x + 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 4\\
x = - 1
\end{array} \right.\\
4)DK:x \ge 1\\
{x^2} - 1 + \sqrt {x - 1} = 0\\
\to \left( {x - 1} \right)\left( {x + 1} \right) + \sqrt {x - 1} = 0\\
\to \sqrt {x - 1} \left( {\sqrt {x - 1} \left( {x + 1} \right) + 1} \right) = 0\\
\to \left[ \begin{array}{l}
x - 1 = 0\\
\sqrt {x - 1} \left( {x + 1} \right) + 1 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
\sqrt {x - 1} \left( {x + 1} \right) + 1 > 0\forall x \ge 1
\end{array} \right.\\
\to x = 1
\end{array}\)
