問題文全文(内容文)：
◎次の不等式を解こう。
$2^{x}-32 \gt 0$
$(\displaystyle \frac{1}{3})^{x-1} \leqq \displaystyle \frac{1}{27}$
$(\displaystyle \frac{1}{4})^{x} \leqq 2^{x+2}$
$16^{x}-3・4^{x}-4 \leqq 0$
$(\displaystyle \frac{1}{3})^{2x-1}+5・(\displaystyle \frac{1}{3})^{x}-2 \lt 0$
◎次の不等式を解こう。
$2^{x}-32 \gt 0$
$(\displaystyle \frac{1}{3})^{x-1} \leqq \displaystyle \frac{1}{27}$
$(\displaystyle \frac{1}{4})^{x} \leqq 2^{x+2}$
$16^{x}-3・4^{x}-4 \leqq 0$
$(\displaystyle \frac{1}{3})^{2x-1}+5・(\displaystyle \frac{1}{3})^{x}-2 \lt 0$
単元：
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指導講師：
とある男が授業をしてみた
問題文全文(内容文)：
◎次の不等式を解こう。
$2^{x}-32 \gt 0$
$(\displaystyle \frac{1}{3})^{x-1} \leqq \displaystyle \frac{1}{27}$
$(\displaystyle \frac{1}{4})^{x} \leqq 2^{x+2}$
$16^{x}-3・4^{x}-4 \leqq 0$
$(\displaystyle \frac{1}{3})^{2x-1}+5・(\displaystyle \frac{1}{3})^{x}-2 \lt 0$
◎次の不等式を解こう。
$2^{x}-32 \gt 0$
$(\displaystyle \frac{1}{3})^{x-1} \leqq \displaystyle \frac{1}{27}$
$(\displaystyle \frac{1}{4})^{x} \leqq 2^{x+2}$
$16^{x}-3・4^{x}-4 \leqq 0$
$(\displaystyle \frac{1}{3})^{2x-1}+5・(\displaystyle \frac{1}{3})^{x}-2 \lt 0$
投稿日：2015.09.16