Đáp án:
\(\begin{array}{l}
3)x = \dfrac{9}{2}\\
4)x = 3\\
5)x \in \emptyset
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
3)\sqrt {{{\left( {x + 3} \right)}^2}} = 3x - 6\\
\to \left| {x + 3} \right| = 3x - 6\\
\to \left[ \begin{array}{l}
x + 3 = 3x - 6\left( {DK:x \ge - 3} \right)\\
x + 3 = - 3x + 6\left( {DK:x < - 3} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
2x = 9\\
4x = 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{9}{2}\\
x = \dfrac{3}{4}\left( l \right)
\end{array} \right.\\
\to x = \dfrac{9}{2}\\
4)\sqrt {{{\left( {x - 2} \right)}^2}} = 2x - 5\\
\to \left| {x - 2} \right| = 2x - 5\\
\to \left[ \begin{array}{l}
x - 2 = 2x - 5\left( {DK:x \ge 2} \right)\\
x - 2 = - 2x + 5\left( {DK:x < 2} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
3x = 7
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
x = \dfrac{7}{3}\left( l \right)
\end{array} \right.\\
\to x = 3\\
5)\sqrt {{{\left( {x - 1} \right)}^2}} + \sqrt {{{\left( {x - 3} \right)}^2}} = 1\\
\to \left| {x - 1} \right| + \left| {x - 3} \right| = 1\\
\to \left[ \begin{array}{l}
x - 1 + x - 3 = 1\left( {DK:x \ge 3} \right)\\
x - 1 - x + 3 = 1\left( {DK:3 > x \ge 1} \right)\\
- x + 1 - x + 3 = 1\left( {DK:1 > x} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
2x = 5\\
2 = 1\left( l \right)\\
- 2x = - 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{5}{2}\left( l \right)\\
x = \dfrac{3}{2}\left( l \right)
\end{array} \right.\\
\to x \in \emptyset
\end{array}\)
