Question

Решить неравенство. Подробно

Answer 1

$7log_9(x^2-3x+2) \leq 8+log_9\frac{(x-2)^7}{x-1}\; ,\\\\ODZ:\; \left \{ {{(x^2-3x+2\ \textgreater \ 0} \atop {\frac{(x-2)^7}{x-1}\ \textgreater \ 0}} \right. ,\; x\in (-\infty,1)U(2,+\infty)\\\\log_9(x-1)^7(x-2)^7 -log_9\frac{(x-2)^7}{x-1}\leq 8\\\\log_9\frac{(x-1)^8(x-2)^7}{(x-2)^7} \leq 8\\\\(x-1)^8 \leq 9^8\\\\P.S.\; a^8-b^8=(a^4-b^4)(a^4+b^4)=\\\\=(a^2-b^2)(a^2+b^2)(a^4+b^4)=(a-b)(a+b)(a^2+b^2)(a^4+b^4)\\\\(x-1-9)(x-1+9)((x-1)^2+9^2)((x-1)^4+9^4) \leq 0$

$(x-10)(x+8) \leq 0\; \; (drygie\; skobki\; \ \textgreater \ 0)\\\\+++(-8)---(10)+++\\\\x\in [\, -8,10\, ]\\\\s \; ychetom\; ODZ:\; \; x\in [-8,1)U(2,10\, ]$