Tìm x biết

a) \(\sqrt {{{\left( {x - 3} \right)}^2}} = 9\).

b) \(\sqrt {4{{\rm{x}}^2} + 4{\rm{x}} + 1} = 6\).

Lời giải

a) \(\sqrt {{{\left( {x - 3} \right)}^2}} = 9\)

\( \Leftrightarrow \left| {x - 3} \right| = 9\)

\( \Leftrightarrow \left[ \begin{array}{l}x - 3 = 9\\x - 3 = - 9\end{array} \right.\)

\( \Leftrightarrow \left[ \begin{array}{l}x = 12\\x = - 6\end{array} \right.\)

Vậy x ∈ {12; – 6}.

b) \(\sqrt {4{{\rm{x}}^2} + 4{\rm{x}} + 1} = 6\)

\( \Leftrightarrow \sqrt {{{\left( {2{\rm{x}} + 1} \right)}^2}} = 6\)

\( \Leftrightarrow \left| {2x + 1} \right| = 6\)

\( \Leftrightarrow \left[ \begin{array}{l}2x + 1 = 6\\2x + 1 = - 6\end{array} \right.\)

\( \Leftrightarrow \left[ \begin{array}{l}2x = 5\\2x = - 7\end{array} \right.\)

\( \Leftrightarrow \left[ \begin{array}{l}x = \frac{5}{2}\\x = \frac{{ - 7}}{2}\end{array} \right.\)

Vậy \[{\rm{x}} \in \left\{ {\frac{5}{2};\frac{{ - 7}}{2}} \right\}\].