Áp dụng bđi côsi, ta có:
$8.2^x=(3+\sqrt 5)^x+(3- \sqrt5)^x+6.(3- \sqrt 5)^x\geq 2\sqrt{(3+ \sqrt5)^x.(3- \sqrt5)^x}+6.(3-\sqrt5)^x=2 \sqrt{4^x}+6.(3-\sqrt5)^x=2.2^x+6.(3-\sqrt5)^x$
$6.2^x\geq 6. (3- \sqrt5)^x\Rightarrow 2^x \geq (3- \sqrt5)^x$
Dấu $=$ xảy ra khi $x=0$
Vậy $x=0$