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Question

$5.2^{2x-1} -7.2^{x-1} +\sqrt{1- 2^{x+2} + 4^{x+1}} =0$

$\log_{2}(\frac{x+1}{2})^{2} -\log_{3}(\frac{x+1}{4})^{3} \leq 2$

score 1 · Answer 1

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

$x=0 , 1-log_210$

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