nếu coi $\sqrt 2 + 1 = t $ thì $\sqrt 2 -1 =\dfrac{1}{t}$ vì $(\sqrt 2 +1)(\sqrt 2 -1) = 1$Vậy ta có $t^{\frac{6x+6}{x+1}} \le \dfrac{1}{t^{-x}}$$\Leftrightarrow t^{\frac{6x+6}{x+1}-x} \le 1 =t^0$$\Leftrightarrow \frac{6x+6}{x+1}-x \le0$
nếu coi $\sqrt 2 + 1 = t $ thì $\sqrt 2 -1 =\dfrac{1}{t}$ vì $(\sqrt 2 +1)(\sqrt 2 -1) = 1$Vậy ta có $t^{\frac{6x+6}{x+1}} \le \dfrac{1}{t^{-x}}$$\Leftrightarrow t^{\frac{6x+6}{x+1}-x} \le 1 =t^0$$\Leftrightarrow \frac{6x+6}{x+1}-x <0$
nếu coi $\sqrt 2 + 1 = t $ thì $\sqrt 2 -1 =\dfrac{1}{t}$ vì $(\sqrt 2 +1)(\sqrt 2 -1) = 1$Vậy ta có $t^{\frac{6x+6}{x+1}} \le \dfrac{1}{t^{-x}}$$\Leftrightarrow t^{\frac{6x+6}{x+1}-x} \le 1 =t^0$$\Leftrightarrow \frac{6x+6}{x+1}-x
\l
e0$