Đáp án:
B6:
a. \(0 \le x < 1\)
Giải thích các bước giải:
\(\begin{array}{l}
B4:\\
a.{x^2} = 9\\
\to \left[ \begin{array}{l}
x = 3\\
x = - 3
\end{array} \right.\\
b.{x^2} = \dfrac{{16}}{{25}}\\
\to \left[ \begin{array}{l}
x = \dfrac{4}{5}\\
x = - \dfrac{4}{5}
\end{array} \right.\\
c.2{x^2} = \dfrac{1}{{50}}\\
\to {x^2} = \dfrac{1}{{100}}\\
\to \left[ \begin{array}{l}
x = \dfrac{1}{{10}}\\
x = - \dfrac{1}{{10}}
\end{array} \right.\\
d.7{x^2} = - 21\left( {voly} \right)\\
Do:{x^2} \ge 0\forall x \in R\\
\to x \in \emptyset \\
B5:\\
a.{x^2} = 10\\
\to \left[ \begin{array}{l}
x = \sqrt {10} \\
x = - \sqrt {10}
\end{array} \right.\\
b.2{x^2} = 6\\
\to {x^2} = 3\\
\to \left[ \begin{array}{l}
x = \sqrt 3 \\
x = - \sqrt 3
\end{array} \right.\\
c.{x^2} = 25\\
\to \left[ \begin{array}{l}
x = 5\\
x = - 5
\end{array} \right.\\
d.5{x^2} + 125 = 0\left( {voly} \right)\\
Do:5{x^2} + 125 > 0\forall x \in R\\
\to x \in \emptyset \\
B6:\\
a.\sqrt x < 1\\
\to 0 \le x < 1\\
b.{x^2} = a\\
\to \left[ \begin{array}{l}
x = a\\
x = - a
\end{array} \right.
\end{array}\)
