js","1717","static/chunks/1717-e9687bfca52045a6.js","8117","static/chunks/8117-3a5cb76430604345.js","1464","static/chunks/1464-4de8faf93b802138.js","422","static/chunks/422-b1f5ec985dcfd2ef.js","3169","static/chunks/3169-0b225cc6264f1f70.js","5791","static/chunks/5791-b7519a5ea5ac96a7.js","4839","static/chunks/4839-abf7392d0b162b93.js","5322","static/chunks/5322-2dee03e10531d285.js","1493","static/chunks/1493-0e365f06d2d7ff74.js","3910","static/chunks/3910-d35f7a55e63a44d2.js","2661","static/chunks/2661-5a8e341dbac91d3f.js","6986","static/chunks/app/(unauthorized-user)/testing/%5BexamType%5D/%5B...slug%5D/page-3d61388e5ed47f00.js"],""] 46:T58b, $5\cdot 2^{2x+2}-21\cdot 2^{x-1}+1\le0 \quad \Leftrightarrow \quad 5\cdot 4\cdot 2^{2x}-\frac{21\cdot 2^x}{2}+1\le0 \; | :2 \quad \Leftrightarrow \quad 40\cdot 2^{2x}-21\cdot 2^x+2\le0.$

Пусть $\; 2^x=t.$ Тогда неравенство примет вид: $\quad 40t^2-21t+2\le0.$

$40t^2-21t+2=0 \quad \Leftrightarrow \quad \begin{cases} t=\frac{1}{8},\\ t=\frac{2}{5}. \end{cases}$ $40t^2-21t+2\le0 \quad \Leftrightarrow \quad (8t-1)(5t-2)\le0.$

Решим полученное неравенство методом интервалов:

$\frac{1}{8}\le t\le \frac{2}{5} \quad \Leftrightarrow \quad 2^{-3}\le 2^x \le 2^{\log_2 \frac{2}{5}} \quad \Leftrightarrow \quad -3\le x \le \log_2 \frac{2}{5}.$

Таким образом, решением исходного неравенства является: $\quad x\in[-3;\; \log_2 \tfrac{2}{5}].$

Ответ: $\quad \left[-3;\; \log_2 \frac{2}{</article></body></html>$

Задание №29388 ЕГЭ по Математике