Решите неравенство

Решите задачу:

$27^{x}-9^{x+1}+ \frac{9^{x+1}-486}{3^{x}-6} \leq 81\; ,\; \; ODZ:\; 3^{x}\ne 6\; ,\; x\ne log_36\\\\(3^{x})^3-(3^{x})^2\cdot 9+ \frac{(3^{x})^2\cdot 9-486}{3^{x}-6} -81 \leq 0\\\\t=3^{x}\; ,\; \; t^3-9t^2+ \frac{9t^2-486}{t-6}-81 \leq 0\\\\ \frac{(t^3-9t^2)(t-6)+9t^2-486-81(t-6)}{t-6} \leq 0\\\\ \frac{t^4-15t^3+63t^2-81t}{t-6} \leq 0\\\\ \frac{t\cdot (t^3-15t^2+63t-81)}{t-6} \leq 0\\\\Pri\; \; t=3:\; \; (t^3-15t^2+63t-81)_{t=3}=0\; \; \to$

$t^3-15t^2+63t-81=(t-3)(t^2-12t+27)=\\\\=(t-3)(t-3)(t-9)=(t-3)^2(t-9)$

$\frac{t\cdot (t-3)^2(t-9)}{t-6} \leq 0\\\\Znaki:\; \; ---[0]+++[3]+++(6)---[9]+++ \\\\t\in (-\infty ,0\, ]\cup (6,9\, ]\; \; \Rightarrow \; \; \left [ {{3^{x} \leq 0} \atop {6\ \textless \ 3^{x} \leq 9}} \right. \; \left [ {{net\; kornej} \atop {3^{log_36}\ \textless \ 3^{x} \leq 3^2}} \right. \\\\log_36\ \textless \ x \leq 2\\\\Otvet:\; \; x\in (log_36\, ;\, 2\, ]\, .$

$3^{3x}-3^{2x+2}+ \frac{3^{2x+2}-486}{3^x-6}-81 \leq 0 \\ \\ (3^x)^3 -3^2*(3^x)^2+ \frac{3^2*(3^x)^2-486}{3^x-6}-81 \leq 0 \\ \\ y=3^x \\ y^3-9y^2+ \frac{9y^2-486}{y-6}-81 \leq 0 \\ \frac{y^3(y-6)-9y^2(y-6)+9y^2-486-81(y-6)}{y-6} \leq 0 \\ \frac{y^4-6y^3-9y^3+54y^2+9y^2-486-81y+486}{y-} \leq 0 \\ \frac{y^4-15y^3+63y^2-81y}{y-6} \leq 0$

ОДЗ: y≠6
(y-6)(y⁴-15y³+63y²-81y)≤0
y(y-6)(y³-15y²+63y-81)≤0

Разложим на множители y³-15y²+63y-81:
при у=9 9³ - 15*9² +63*9 -81=729-1215+567-81=0

Разделим y³-15y²+63y-81 на (у-9):
y³ -15y²+63y-81 | y-9
- ------------
y³ - 9y² y² -6y +9
--------------
_ -6y² + 63y
-6y² + 54y
----------------
_ 9y - 81
9y - 81
------------
0

y³-15y²+63y-81=(y-9)(y²-6y+9)=(y-9)(y-3)²

y(y-3)²(y-6)(y-9)≤0
y=0 y=3 y=6 y=9
- + + - +
------- 0 ----------- 3 ------------ 6 ------------ 9 ------------
\\\\\\\\\ \\\ \\\\\\\\\\\\\\\\
y≤0
y=3
6
3ˣ ≤ 0
нет решений

3ˣ=3
х=1

6< 3ˣ ≤9
3ˣ>6
x>log₃ 6

3ˣ ≤ 9
3ˣ ≤ 3²
x ≤ 2

x∈{1}U(log₃ 6; 2]