Решить неравенство: log2(x^2 - 3x + 2) ≤ log2(x-2) + 1

Question 1

Решить неравенство:
log2(x^2 - 3x + 2) ≤ log2(x-2) + 1

Question 2

$log_2(x^2 - 3x + 2) \leq log_2(x-2) + 1$
ОДЗ:
$\left \{ {{ x^{2} -3x+2\ \textgreater \ 0} \atop {x-2\ \textgreater \ 0}} \right.$
$\left \{ {{ (x-2)(x-1)\ \textgreater \ 0} \atop {x\ \textgreater \ 2}} \right.$

------+-----(1)------- - --------(2)------+-------
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----------------------------------(2)--------------
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$x$ ∈ $(2;+$ ∞ $)$

$log_2(x^2 - 3x + 2) \leq log_2(x-2) + log_22$
$log_2(x^2 - 3x + 2) \leq log_2[2(x-2)]$
$x^2 - 3x + 2 \leq 2(x-2)$
$x^2 - 3x + 2 \leq 2x-4$
$x^2 - 3x + 2- 2x+4 \leq 0$
$x^2 - 5x + 6 \leq 0$
$D=(-5)^2-4*1*6=1$
$x_1= \frac{5+1}{2} =3$
$x_2= \frac{5-1}{2} =2$
$(x-2)(x-3) \leq 0$

-------+--------[2]----- - -----[3]-------+-------
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Ответ: $(2;3]$

Question 3

Спасибо огромное!

Luluput_zn · Answer 1 · 2018-05-11T08:26:55+0000

$log_2(x^2 - 3x + 2) \leq log_2(x-2) + 1$
ОДЗ:
$\left \{ {{ x^{2} -3x+2\ \textgreater \ 0} \atop {x-2\ \textgreater \ 0}} \right.$
$\left \{ {{ (x-2)(x-1)\ \textgreater \ 0} \atop {x\ \textgreater \ 2}} \right.$

------+-----(1)------- - --------(2)------+-------
////////////// //////////////////
----------------------------------(2)--------------
///////////////////
$x$ ∈ $(2;+$ ∞ $)$

$log_2(x^2 - 3x + 2) \leq log_2(x-2) + log_22$
$log_2(x^2 - 3x + 2) \leq log_2[2(x-2)]$
$x^2 - 3x + 2 \leq 2(x-2)$
$x^2 - 3x + 2 \leq 2x-4$
$x^2 - 3x + 2- 2x+4 \leq 0$
$x^2 - 5x + 6 \leq 0$
$D=(-5)^2-4*1*6=1$
$x_1= \frac{5+1}{2} =3$
$x_2= \frac{5-1}{2} =2$
$(x-2)(x-3) \leq 0$

-------+--------[2]----- - -----[3]-------+-------
//////////////////
-----------------(2)------------------------------
/////////////////////////////////

Ответ: $(2;3]$