Mn giúp mk lm bài 9 với cảm ơn ạ Question Mn giúp mk lm bài 9 với cảm ơn ạ in progress 0 Môn Toán Nem 4 years 2020-10-29T01:45:46+00:00 2020-10-29T01:45:46+00:00 1 Answers 79 views 0
Answers ( )
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\sqrt x = 8\,\,\,\,\,\left( {x \ge 0} \right)\\
\Leftrightarrow x = {8^2}\\
\Leftrightarrow x = 64\\
b,\\
3\sqrt x = 12\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge 0} \right)\\
\Leftrightarrow \sqrt x = 4\\
\Leftrightarrow x = {4^2}\\
\Leftrightarrow x = 16\\
c,\\
5\sqrt {2x} = 30\,\,\,\,\,\,\,\left( {x \ge 0} \right)\\
\Leftrightarrow \sqrt {2x} = 6\\
\Leftrightarrow 2x = {6^2}\\
\Leftrightarrow 2x = 36\\
\Leftrightarrow x = 18\\
d,\\
\sqrt x < 3\,\,\,\,\,\,\,\,\,\,\left( {x \ge 0} \right)\\
\Leftrightarrow 0 \le x < 9\\
e,\\
\sqrt x \ge 5\,\,\,\,\,\,\,\,\,\left( {x \ge 0} \right)\\
\Leftrightarrow x \ge {5^2}\\
\Leftrightarrow x \ge 25\\
f,\\
\sqrt {2x} \le 6\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge 0} \right)\\
\Leftrightarrow 2x \le {6^2}\\
\Leftrightarrow 2x \le 36\\
\Leftrightarrow 0 \le x \le 18
\end{array}\)