Giải thích các bước giải:
Ta có:
$\left(\sqrt{10}-3\right)^{\frac{3-x}{x-1}}<\left(\sqrt{10}+3\right)^{\frac{x+1}{x+3}}$
$\to \left(\dfrac{\left(\sqrt{10}-3\right)\left(\sqrt{10}+3\right)}{\sqrt{10}+3}\right)^{\frac{3-x}{x-1}}<\left(\sqrt{10}+3\right)^{\frac{x+1}{x+3}}$
$\to \left(\dfrac{10-3^2}{\sqrt{10}+3}\right)^{\frac{3-x}{x-1}}<\left(\sqrt{10}+3\right)^{\frac{x+1}{x+3}}$
$\to \left(\dfrac{1}{\sqrt{10}+3}\right)^{\frac{3-x}{x-1}}<\left(\sqrt{10}+3\right)^{\frac{x+1}{x+3}}$
$\to \dfrac{1}{\left(\sqrt{10}+3\right)^{\frac{3-x}{x-1}}}<\left(\sqrt{10}+3\right)^{\frac{x+1}{x+3}}$
$\to \left(\sqrt{10}+3\right)^{\frac{3-x}{x-1}}\cdot \left(\sqrt{10}+3\right)^{\frac{x+1}{x+3}}>1$
$\to \left(\sqrt{10}+3\right)^{\frac{3-x}{x-1}+\frac{x+1}{x+3}}>1$
$\to \left(\sqrt{10}+3\right)^{\frac{8}{\left(x-1\right)\left(x+3\right)}}>1$
$\to \dfrac{8}{\left(x-1\right)\left(x+3\right)}>0$
$\to \left(x-1\right)\left(x+3\right)>0$
$\to x>1$ hoặc $x<-3$
