Giải các phương trình:
a) \(\sqrt {3x - 4} = x - 3\)
b) \(\sqrt {{x^2} - 2x + 3} = 2x - 1\)
c) \(\sqrt {2{x^2} + 3x + 7} = x + 2\)
d) \(\sqrt {3{x^2} - 4x - 4} = \sqrt {2x +5} \)
a) \(\sqrt {3x - 4} = x - 3\)
\(\begin{array}{l}
\Leftrightarrow \left\{ \begin{array}{l}
x - 3 \ge 0\\
3x - 4 = {\left( {x - 3} \right)^2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 3\\
{x^2} - 9x + 13 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 3\\
x = \frac{{9 \pm \sqrt {29} }}{2}
\end{array} \right. \Leftrightarrow x = \frac{{9 + \sqrt {29} }}{2}
\end{array}\)
Vậy phương trình có 1 nghiệm \(x = \frac{{9 + \sqrt {29} }}{2}\)
b) \(\sqrt {{x^2} - 2x + 3} = 2x - 1\)
\(\begin{array}{l}
\Leftrightarrow \left\{ \begin{array}{l}
2x - 1 \ge 0\\
{x^2} - 2x + 3 = {\left( {2x - 1} \right)^2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge \frac{1}{2}\\
3{x^2} - 2x - 2 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge \frac{1}{2}\\
x = \frac{{1 \pm \sqrt 7 }}{3}
\end{array} \right. \Leftrightarrow x = \frac{{1 + \sqrt 7 }}{3}
\end{array}\)
Vậy phương trình có 1 nghiệm \(x = \frac{{1 + \sqrt 7 }}{3}\)
c) \(\sqrt {2{x^2} + 3x + 7} = x + 2\)
\(\begin{array}{l}
\Leftrightarrow \left\{ \begin{array}{l}
x > - 2\\
2{x^2} + 3x + 7 = {x^2} + 4x + 4
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x > - 2\\
{x^2} - x + 3 = 0\,\,\left( {VN} \right)
\end{array} \right.
\end{array}\)
Vậy phương trình vô nghiệm
d) \(\sqrt {3{x^2} - 4x - 4} = \sqrt {2x + 5} \)
\(\begin{array}{l}
\Leftrightarrow \left\{ \begin{array}{l}
x \ge - \frac{5}{2}\\
3{x^2} - 4x - 4 = 2x + 5
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge - \frac{5}{2}\\
3{x^2} - 6x - 9 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = - 1
\end{array} \right.
\end{array}\)
-- Mod Toán 10