Question

Решите неравенство нужно полное решение

Answer 1

$7\, log_9(x^2+3x-10)\leq 8+log_9\dfrac{(x-2)^7}{x+5}\ \ ,\\\\\\ODZ:\ \left\{\begin{array}{l}x^2+3x-10>0\\\frac{(x-2)^7}{x+5}>0\end{array}\right\ \ ,\ \ \left\{\begin{array}{ccc}(x-2)(x+5)>0\\(x-2)^7(x+5)>0\end{array}\right\ ,\\\\x\in (-\infty ;-5\, )\cup (\, 2;+\infty )\\\\ log_9(x^2+3x-10)\leq 8+log_9\dfrac{(x-2)^7}{x+5}\\\\\\ log_9(x-2)^7(x+5)^7-log_9\dfrac{(x-2)^7}{x+5}\leq 8\\\\\\log_9\dfrac{(x-2)^7(x+5)^7}{\frac{(x-2)^7}{x+5}}\leq log_99^8\ \ ,\ \ \ \ \dfrac{(x-2)^7(x+5)^8}{(x-2)^7}\leq 9^8\ \ ,$

$(x+5)^8-9^8\leq 0\ \ ,\\\\\star \ \ a^8-b^8=(a^4-b^4)(a^4+b^4)=(a-b)(a+b)(a^2+b^2)(a^4+b^4)\ \ \star \\\\(x+5-9)(x+5+9)(\underbrace {(x+5)^2+81}_{> 0})(\underbrace {(x+5)^4+6561}_{>0})\leq 0\\\\(x-4)(x+14)\leq 0\ \ \ \ +++(-14)---(4)+++\\\\x\in [-14\, ;\, 4\ ]\ \ \ ,\ \ \ \left\{\begin{array}{l}x\in (-\infty \ ;\ -5)\cup (\ 2\, ;+\infty \, )\\x\in [-14\, ;\, 4\ ]\end{array}\right\ \ \ \Rightarrow \\\\\\\underline {\ x\in [-14\, ;\, -5)\cup (\ 2\, ;\, 4\ )\ }$