Đáp án:
\(\begin{array}{l}
6)\left[ \begin{array}{l}
x = 3\\
x = 0
\end{array} \right.\\
7)\left[ \begin{array}{l}
x = \dfrac{2}{3}\\
x = - 4
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
6)\sqrt {{{\left( {x - 1} \right)}^2}} + \sqrt {{{\left( {x - 2} \right)}^2}} = 3\\
\to \left| {x - 1} \right| + \left| {x - 2} \right| = 3\\
\to \left[ \begin{array}{l}
x - 1 + x - 2 = 3\left( {DK:x \ge 2} \right)\\
x - 1 - x + 2 = 3\left( {2 > x \ge 1} \right)\\
- x + 1 - x + 2 = 3\left( {1 > x} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
2x = 6\\
1 = 3\left( l \right)\\
- 2x = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
x = 0
\end{array} \right.\\
7)\sqrt {{x^2} - 6x + 9} = \sqrt {4{x^2} + 4x + 1} \\
\to {x^2} - 6x + 9 = 4{x^2} + 4x + 1\\
\to 3{x^2} + 10x - 8 = 0\\
\to \left( {3x - 2} \right)\left( {x + 4} \right) = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{2}{3}\\
x = - 4
\end{array} \right.
\end{array}\)
