Bài 1.$5.2^{2x-1}-7.2^{x-1}+\sqrt{1-2^{x+2}+4^{x+1}}=0$
Ta có
$1-2^{x+2}-4^{x+1}=(1-2^{x+1})^2$
$\Rightarrow \sqrt{1-2^{x+2}+4^{x+1}}=1-2^{x+1}$ khi $x\le -1$ , $=2^{x+1}-1$ khi $x\ge -1$
TH1: $x\ge -1$
$5.2^{2x-1}-7.2^{x-1}+2^{x+1}-1=0$
Đặt $2^{x-1}=a>0\Rightarrow 2^{2x-1}=2a^2 , 2^{x+1}=4a$
$10a^2-7a+4a-1=0$
$\Rightarrow 10a^2-3a-1=0$
$\Rightarrow a=\frac{1}{2} $
$\Rightarrow x-1=-1\Rightarrow x=0$ Thỏa mãn
TH2. $x\le -1$
$5.2^{2x-1}-7.2^{x-1}+1-2^{x+1}=0$
$\Rightarrow 10a^2-7a+1-4a=0$
$\Rightarrow 10a^2-11a+1=0$
$\Rightarrow a=1, \frac{1}{10}$
$\Rightarrow x-1=0 , -log_210$
$\Rightarrow x=1 , 1-log_210$
Vì $x\le -1 \Rightarrow x=1-log_210$
Vậy $x=0 , 1-log_210$