Đáp án:
n) \(1 > x \ge \dfrac{2}{3}\)
Giải thích các bước giải:
\(\begin{array}{l}
g)\sqrt {{x^2} - 6x + 9} = 2x - 1\\
\to \sqrt {{{\left( {x - 3} \right)}^2}} = 2x - 1\\
\to \left| {x - 3} \right| = 2x - 1\\
\to \left[ \begin{array}{l}
x - 3 = 2x - 1\left( {DK:x \ge 3} \right)\\
x - 3 = - 2x + 1\left( {DK:x < 3} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 2\left( l \right)\\
3x = 4
\end{array} \right.\\
\to x = \dfrac{4}{3}\\
h)\sqrt {{x^2} + 6x + 9} = 5\\
\to \sqrt {{{\left( {x + 3} \right)}^2}} = 5\\
\to \left| {x + 3} \right| = 5\\
\to \left[ \begin{array}{l}
x + 3 = 5\\
x + 3 = - 5
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\\
x = - 8
\end{array} \right.\\
i)\sqrt {{x^2} - 4x + 4} = 2\\
\to \sqrt {{{\left( {x - 2} \right)}^2}} = 2\\
\to \left| {x - 2} \right| = 2\\
\to \left[ \begin{array}{l}
x - 2 = 2\\
x - 2 = - 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 4\\
x = 0
\end{array} \right.\\
n)DK:x \ge \dfrac{2}{3}\\
\to \sqrt {3x - 2} < 1\\
\to 3x - 2 < 1\\
\to 3x < 3\\
\to x < 1\\
\to 1 > x \ge \dfrac{2}{3}
\end{array}\)
\(\begin{array}{l}
f)\sqrt {4{x^2} + 4x + 1} = 2 + x\\
\to \sqrt {{{\left( {2x + 1} \right)}^2}} = 2 + x\\
\to \left| {2x + 1} \right| = x + 2\\
\to \left[ \begin{array}{l}
2x + 1 = x + 2\\
2x + 1 = - x - 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
x = - 1
\end{array} \right.
\end{array}\)
