решите два задания(как можно подробнее). Created by vityamath. algebra-ru

решите два задания(как можно подробнее)...

mmb1

Verified answer

(lg 300 - lg 3) / (3log(6) 2 + log(6) 27) = lg (300/3) / (log(6) 2³ + log(6) 3³) = lg 100 / log(6) 6³ = 2/3

log²(2) x + 4log(2) x - 5 ≥ 0

x > 0

log(2) x = y

y² + 4y - 5 ≥ 0

(y + 5)(y - 1) ≥ 0

+++++++[-5] ------------- [1] +++++++++

1. y ≤ -5

log(2) x ≤ 5

x ≤ 2^-5 = 1/32

2/ y ≥ 1

log(2) x ≥ 1

x ≥ 2

x ∈ (0, 1/32] U [2, +∞)

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bb573878

Ответ:

Объяснение:

$\displaystyle\bf\\\frac{lg300-lg3}{3log_62+log_627} =\frac{lg\dfrac{300}{3} }{log_62^3+log_63^3} =\frac{lg100}{log_6(2^3\cdot3^3)}=\frac{lg10^2}{log_66^3}=\frac{2}{3}$

$\displaystyle\bf\\log_2^2x+4log_2x-5\geq 0;\ \ \ \ \ \ \ \ ODZ:x>0\\\\log_2^2x+5log_2x-log_2x-5\geq 0\\\\log_2x(log_2x+5)-(log_2x+5)\geq 0\\\\(log_2x+5)(log_2x-1)\geq 0\\\\(log_2x-log_22^{-5})(log_2x-log_22)\geq 0\\\\\bigg(x-\frac{1}{32} \bigg)\bigg(x-2\bigg)\geq 0\\\\+++\bigg[\frac{1}{32} \bigg]---\bigg[2\bigg]+++>x\\\left \{ {{x>0} \atop {x\in\bigg(-\infty;\dfrac{1}{32} \bigg];\bigg[2;+\infty\bigg)}} \right. \\\\\\Otvet:x\in\bigg(0;\dfrac{1}{32} \bigg];\bigg[2;+\infty\bigg)}$

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