Đáp án:
\(\begin{array}{l}
a)\left[ \begin{array}{l}
x = \sqrt 3 \\
x = - \sqrt 3
\end{array} \right.\\
b)x = \sqrt 5 \\
c)\left[ \begin{array}{l}
x = \dfrac{1}{{\sqrt 2 }}\\
x = - \dfrac{1}{{\sqrt 2 }}
\end{array} \right.\\
d)x = \dfrac{1}{{\sqrt 2 }}\\
h)x = \dfrac{3}{2}\\
d)\left[ \begin{array}{l}
x = 8\\
x = - 8
\end{array} \right.\\
e)\left[ \begin{array}{l}
x = \dfrac{{19}}{2}\\
x = - \dfrac{{19}}{2}
\end{array} \right.\\
f)\left[ \begin{array}{l}
x = 2\\
x = - 2
\end{array} \right.\\
g)\left[ \begin{array}{l}
x = 5\\
x = - 1
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a){x^2} = 3\\
\to \left[ \begin{array}{l}
x = \sqrt 3 \\
x = - \sqrt 3
\end{array} \right.\\
b){x^2} - 2.x.\sqrt 5 + 5 = 0\\
\to {\left( {x - \sqrt 5 } \right)^2} = 0\\
\to x - \sqrt 5 = 0\\
\to x = \sqrt 5 \\
c)4{x^2} = 2\\
\to {x^2} = \dfrac{1}{2}\\
\to \left[ \begin{array}{l}
x = \dfrac{1}{{\sqrt 2 }}\\
x = - \dfrac{1}{{\sqrt 2 }}
\end{array} \right.\\
d)4{x^2} - 2.2x.\sqrt 2 + 2 = 0\\
\to {\left( {2x - \sqrt 2 } \right)^2} = 0\\
\to 2x - \sqrt 2 = 0\\
\to x = \dfrac{1}{{\sqrt 2 }}\\
h)\sqrt {{{\left( {x - 2} \right)}^2}} = 1 - x\\
\to \left| {x - 2} \right| = 1 - x\\
\to \left[ \begin{array}{l}
x - 2 = 1 - x\\
x - 2 = - 1 + x
\end{array} \right.\\
\to \left[ \begin{array}{l}
2x = 3\\
- 2 = - 1\left( l \right)
\end{array} \right.\\
\to x = \dfrac{3}{2}\\
d)\sqrt {{x^2}} = 8\\
\to \left| x \right| = 8\\
\to \left[ \begin{array}{l}
x = 8\\
x = - 8
\end{array} \right.\\
e)2\left| x \right| = 19\\
\to \left| x \right| = \dfrac{{19}}{2}\\
\to \left[ \begin{array}{l}
x = \dfrac{{19}}{2}\\
x = - \dfrac{{19}}{2}
\end{array} \right.\\
f)7\left| x \right| = 17\\
\to \left| x \right| = 2\\
\to \left[ \begin{array}{l}
x = 2\\
x = - 2
\end{array} \right.\\
g)\left| {x - 2} \right| = 3\\
\to \left[ \begin{array}{l}
x - 2 = 3\\
x - 2 = - 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 5\\
x = - 1
\end{array} \right.
\end{array}\)
