Đáp án:
\(\begin{array}{l}
d,\\
Phương\,\,trình\,\,vô\,\,nghiệm\\
e,\\
Phương\,\,trình\,\,vô\,\,nghiệm\\
f,\\
x = \dfrac{7}{2}
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
d,\\
DKXD:\,\,\,{x^2} + 4x + 4 \ge 0 \Leftrightarrow {\left( {x + 2} \right)^2} \ge 0,\,\,\,\forall x\\
\sqrt {{x^2} + 4x + 4} = x - 3\\
\Leftrightarrow \sqrt {{x^2} + 2.x.2 + {2^2}} = x - 3\\
\Leftrightarrow \sqrt {{{\left( {x + 2} \right)}^2}} = x - 3\\
\Leftrightarrow \left| {x + 2} \right| = x - 3\\
\Leftrightarrow \left\{ \begin{array}{l}
x - 3 \ge 0\\
\left[ \begin{array}{l}
x + 2 = x - 3\\
x + 2 = - \left( {x - 3} \right)
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 3\\
\left[ \begin{array}{l}
x + 2 - x + 3 = 0\\
x + 2 + x - 3 = 0
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 3\\
\left[ \begin{array}{l}
5 = 0\,\,\,\,\left( L \right)\\
2x - 1 = 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 3\\
x = \dfrac{1}{2}
\end{array} \right.\\
\Rightarrow Phương\,\,trình\,\,vô\,\,nghiệm\\
e,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
2x - 3 \ge 0\\
5x - 6 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{3}{2}\\
x \ge \dfrac{6}{5}
\end{array} \right. \Leftrightarrow x \ge \dfrac{3}{2}\\
\sqrt {2x - 3} = \sqrt {5x - 6} \\
\Leftrightarrow {\sqrt {2x - 3} ^2} = {\sqrt {5x - 6} ^2}\\
\Leftrightarrow 2x - 3 = 5x - 6\\
\Leftrightarrow 2x - 3 - 5x + 6 = 0\\
\Leftrightarrow - 3x + 3 = 0\\
\Leftrightarrow 3x = 3\\
\Leftrightarrow x = 1\\
x \ge \dfrac{3}{2} \Rightarrow Phương\,\,trình\,\,vô\,\,nghiệm\\
f,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
2x - 1 \ge 0\\
x - 2 \ge 0\\
\sqrt {x - 2} \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
2x - 1 \ge 0\\
x - 2 > 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{1}{2}\\
x > 2
\end{array} \right. \Leftrightarrow x > 2\\
\dfrac{{\sqrt {2x - 1} }}{{\sqrt {x - 2} }} = 2\\
\Leftrightarrow \sqrt {2x - 1} = 2\sqrt {x - 2} \\
\Leftrightarrow {\sqrt {2x - 1} ^2} = {\left( {2\sqrt {x - 2} } \right)^2}\\
\Leftrightarrow 2x - 1 = 4.{\sqrt {x - 2} ^2}\\
\Leftrightarrow 2x - 1 = 4.\left( {x - 2} \right)\\
\Leftrightarrow 2x - 1 = 4x - 8\\
\Leftrightarrow 2x - 1 - 4x + 8 = 0\\
\Leftrightarrow - 2x + 7 = 0\\
\Leftrightarrow 2x = 7\\
\Leftrightarrow x = \dfrac{7}{2}\,\,\,\,\,\left( {t/m} \right)\\
\Rightarrow x = \dfrac{7}{2}
\end{array}\)
