Đáp án:
\(\begin{array}{l}
VD1:\\
a,\,\,13\\
b,\,\,13\\
VD2:\\
a,\,\,3 - \sqrt 2 \\
b,\,\,\,1 + \sqrt 2 \\
c,\,\,\sqrt 5 - 2\\
VD:\\
a,\,\,\,x - 3\\
d,\,\,\, - \left( {x + 3} \right)\\
e,\,\,\,\,7 - x\\
c,\,\,\,\,\left[ \begin{array}{l}
\sqrt {{{\left( {x - 2} \right)}^2}} = x - 2,\,\,\,\,\,x \ge 2\\
\sqrt {{{\left( {x - 2} \right)}^2}} = 2 - x,\,\,\,\,\,x < 2
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
VD1:\\
a,\,\,\sqrt {{{13}^2}} = \left| {13} \right| = 13\\
b,\,\,\sqrt {{{\left( { - 13} \right)}^2}} = \left| { - 13} \right| = 13\\
VD2:\\
a,\,\,\sqrt {{{\left( {\sqrt 2 - 3} \right)}^2}} = \left| {\sqrt 2 - 3} \right| = - \left( {\sqrt 2 - 3} \right) = 3 - \sqrt 2 \\
b,\,\,\,\sqrt {{{\left( {1 + \sqrt 2 } \right)}^2}} = \left| {1 + \sqrt 2 } \right| = 1 + \sqrt 2 \\
c,\,\,\sqrt {{{\left( {\sqrt 5 - 2} \right)}^2}} = \left| {\sqrt 5 - 2} \right| = \sqrt 5 - 2\\
VD:\\
a,\,\,\,x \ge 3 \Rightarrow x - 3 \ge 0 \Rightarrow \left| {x - 3} \right| = x - 3\\
\sqrt {{{\left( {x - 3} \right)}^2}} = \left| {x - 3} \right| = x - 3\\
d,\,\,\,x < - 3 \Leftrightarrow x + 3 < 0 \Leftrightarrow \left| {x + 3} \right| = - \left( {x + 3} \right)\\
\sqrt {{{\left( {x + 3} \right)}^2}} = \left| {x + 3} \right| = - \left( {x + 3} \right)\\
e,\,\,\,\,x < 7 \Rightarrow 7 - x > 0 \Rightarrow \left| {7 - x} \right| = 7 - x\\
\sqrt {{{\left( {7 - x} \right)}^2}} = \left| {7 - x} \right| = 7 - x\\
c,\,\,\,\,\sqrt {{{\left( {x - 2} \right)}^2}} = \left| {x - 2} \right|\\
TH1:\,\,\,x \ge 2 \Rightarrow x - 2 \ge 0 \Rightarrow \left| {x - 2} \right| = x - 2\\
TH2:\,\,\,x < 2 \Rightarrow x - 2 < 0 \Rightarrow \left| {x - 2} \right| = - \left( {x - 2} \right) = 2 - x\\
\Rightarrow \left[ \begin{array}{l}
\sqrt {{{\left( {x - 2} \right)}^2}} = x - 2,\,\,\,\,\,x \ge 2\\
\sqrt {{{\left( {x - 2} \right)}^2}} = 2 - x,\,\,\,\,\,x < 2
\end{array} \right.
\end{array}\)
