Question

помогите пожалуйста решить

Answer 1

Ответ:

$x=0$

$x=3$

$x=n\pi$

Объяснение:

$5^{|x+1|}=(0,2)^{x-1}$

$5^{|x+1|}=(\frac{2}{10})^{x-1}$

$5^{|x+1|}=(\frac{1}{5})^{x-1}$

$5^{|x+1|}=(5^{-1})^{x-1}$

$5^{|x+1|}=5^{-(x-1)}$

$5^{|x+1|}=5^{-x+1}$

$|x+1|=-x+1$

1.

$-x+1\ge0$

$x\le 1$

2.

$x\in(-\infty;0]$

$-(x+1)=-x+1$

$-x-1=-x+1$

$-x+x=1+1$

$0=2$

$x\in \emptyset$

3.

$x\in(0;1]$

$x+1=-x+1$

$x+x=1-1$

$2x=0 \ \ \ |:2$

$\underline{x=0}$

======================

$4^x-2^{2(x-1)}+8^{\frac{2(x-2)}{3}}=52$

$(2^2)^x-2^{2(x-1)}+(2^3)^{\frac{2(x-2)}{3}}=52$

$2^{2x}-2^{2x-2}+2^{3\cdot\frac{2(x-2)}{3}}=52$

$2^{2x}-2^{2x-2}+2^{2(x-2)}=52$

$2^{2x}-2^{2x}\cdot 2^{-2}+2^{2x-4}=52$

$2^{2x}-2^{2x}\cdot 2^{-2}+2^{2x}\cdot 2^{-4}=52$

$2^{2x}-2^{2x}\cdot \frac{1}{4}+2^{2x}\cdot \frac{1}{16}=52\ \ \ |\cdot16$

$16\cdot 2^{2x}-4\cdot 2^{2x}+2^{2x}=832$

$13\cdot2^{2x}=832\ \ \ |:13$

$2^{2x}=64$

$2^{2x}=2^6$

$2x=6\ \ \ |:2$

$\underline{x=3}$

======================

$2^{cos 2x}=3\cdot 2^{cos^2x}-4$

$2^{2cos^2x-1}-3\cdot 2^{cos^2x}+4=0$

$2^{2cos^2x}\cdot 2^{-1}-3\cdot 2^{cos^2x}+4=0$

$2^{2cos^2x}\cdot \frac{1}{2}-3\cdot 2^{cos^2x}+4=0\ \ \ |\cdot 2$

$2^{2cos^2x}-6\cdot 2^{cos^2x}+8=0$

$2^{cos^2x}=t,\ t\in[1;2]$

$t^2-6t+8=0$

$D=(-6)^2-4\cdot1\cdot8=36-32=4$

$\sqrt{D}=\sqrt4=2$

$t_1=\frac{6-2}{2\cdot1}=\frac{4}{2}=2$

$t_2=\frac{6+2}{2\cdot1}=\frac{8}{2}=8\not \in [1;2]$

$2^{cos^2x}=2$

$2^{cos^2x}=2^1$

$cos^2x=1$

$\underline{x=n\pi}$