Đáp án: $ x=1\quad or\quad x=2$
Giải thích các bước giải:
$(\sqrt[]{2}-1)^{x^2-5x+4}=(3+2\sqrt[]{2})^{x-1}$
$\rightarrow (\sqrt[]{2}-1)^{x^2-5x+4}=(2+2\sqrt[]{2}+1)^{x-1}$
$\rightarrow (\sqrt[]{2}-1)^{x^2-5x+4}=((\sqrt[]{2}+1)^2)^{x-1}$
$\rightarrow (\sqrt[]{2}-1)^{x^2-5x+4}=(\sqrt[]{2}+1)^{2x-2}$
$\rightarrow (\sqrt[]{2}-1)^{x^2-5x+4}.(\sqrt[]{2}-1)^{2x-2}=(\sqrt[]{2}+1)^{2x-2}.(\sqrt[]{2}-1)^{2x-2}$
$\rightarrow (\sqrt[]{2}-1)^{x^2-5x+4+2x-2}=((\sqrt[]{2}+1)(\sqrt[]{2}-1))^{2x-2}$
$\rightarrow (\sqrt[]{2}-1)^{x^2-3x+2}=1^{2x-2}$
$\rightarrow (\sqrt[]{2}-1)^{x^2-3x+2}=1$
$\rightarrow x^2-3x+2=0$
$\rightarrow (x-1)(x-2)=0$
$\rightarrow x=1\quad or\quad x=2$
