Đáp án: a.$-\sqrt{5}<x<-2\quad \mathrm{hoặc}\quad \:1<x<\sqrt{5}$
b.$\dfrac12\le x<1$
Giải thích các bước giải:
a.ĐKXD : $-3<x<-2$ hoặc $x>1$
Ta có :
$\log_{0.1}(x^2+x-2)>\log_{0.1}(x+3)$
$\to x^2+x-2<x+3$ vì $0.1<1$
$\to x^2<5$
$\to -\sqrt5<x<\sqrt5$
$\to -\sqrt{5}<x<-2\quad \mathrm{hoặc}\quad \:1<x<\sqrt{5}$
b.ĐKXĐ: $x<1$
Ta có :
$\log_{\dfrac13}(x^2-6x+5)+2\log_3(2-x)\ge 0$
$\to \log_{\dfrac13}(x^2-6x+5)+\log_3(2-x)^2\ge 0$
$\to \log_{\dfrac13}(x^2-6x+5)\ge -\log_3(2-x)^2$
$\to \log_{\dfrac13}(x^2-6x+5)\ge \log_{\dfrac13}(2-x)^2$
$\to x^2-6x+5\le (2-x)^2$ vì $\dfrac13<1$
$\to x^2-6x+5\le x^2-4x+4$
$\to 2x\ge 1$
$\to x\ge \dfrac12$
$\to\dfrac12\le x<1$
