a)

${\left(2\sqrt{2}\right)}^x={\left(\sqrt[3]{4}\right)}^{-2x+3}$  

${\left(2^1\cdot 2^{\frac{1}{2}}\right)}^x={\left(\sqrt[3]{2^2}\right)}^{-2x+3}$      

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

$2^{\frac{3}{2}x}=2^{-\frac{4}{3}x+2}$  

$\frac{3}{2}x=-\frac{4}{3}x+2\mid \cdot 6$  

$9x=-8x+12\mid +8x$  

$17x=12\mid :17$  

$x=\frac{12}{17}$  

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b)

$2^{\frac{2x^2-4x+2}{x^2-4}}=8^{\frac{1}{x-2}}$  

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

$2^{\frac{2x^2-4x+2}{\left(x-2\right)\left(x+2\right)}}=2^{\frac{3}{x-2}}$  

$\frac{2x^2-4x+2}{\left(x-2\right)\left(x+2\right)}=\frac{3}{x-2}\mid \cdot \left(x-2\right)\left(x+2\right)$  

$2x^2-4x+2=3\left(x+2\right)$  

$2x^2-4x+2=3x+6\mid -3x-6$  

$2x^2-7x-4=0$  

$\Delta ={\left(-7\right)}^2-4\cdot 2\cdot \left(-4\right)=49+32=81$  

$x_1=\frac{7-9}{2\cdot 2}=\frac{-2}{4}=-\frac{1}{2}$  

$x_2=\frac{7+9}{2\cdot 2}=\frac{16}{4}=4$  

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c)

$\left(2^{4x+2}-20\cdot 2^{2x+1}+64\right)\left(3^{x-4}+24\cdot 3^{\frac{1}{2}x-2}-81\right)=0$  

$\left({\left(2^{2x+1}\right)}^2-20\cdot 2^{2x+1}+64\right)\left({\left(3^{\frac{1}{2}x-2}\right)}^2+24\cdot 3^{\frac{1}{2}x-2}-81\right)=0$  

Podstawmy  $2^{2x+1}=t,3^{\frac{1}{2}x-2}=p$  

$\left(t^2-20t+64\right)\left(p^2+24p-81\right)=0$  

$t^2-20t+64=0\vee p^2+24p-81=0$  

$\Delta ={\left(-20\right)}^2-4\cdot 1\cdot 64=400-256=144,\Delta ={24}^2-4\cdot 1\cdot \left(-81\right)=576+324=900$  

$t_1=\frac{20-12}{2}=\frac{8}{2}=4,t_2=\frac{20+12}{2}=\frac{32}{2}=16,p_1=\frac{-24-30}{2}=\frac{-74}{2}=-36,p_2=\frac{-24+30}{2}=\frac{6}{2}=3$  

Zatem otrzymujemy:

$2^{2x+1}=4\vee 2^{2x+1}=16\vee 3^{\frac{1}{2}x-2}=-36\vee 3^{\frac{1}{2}x-2}=3$  

$2^{2x+1}=2^2\vee 2^{2x+1}=2^4\vee \text{sprzecznosˊcˊ}\vee 3^{\frac{1}{2}x-2}=3^1$  

$2x+1=2\vee 2x+1=4\vee \frac{1}{2}x-2=1$  

$2x=1\vee 2x=3\vee \frac{1}{2}x=3$  

$x=\frac{1}{2}\vee x=\frac{3}{2}\vee x=6$  

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d)

$5\cdot 3^{|10-2x|}-\frac{50}{3}\cdot 3^{|5-x|}=-5$  

$5\cdot 3^{2|5-x|}-\frac{50}{3}\cdot 3^{|5-x|}=-5$  

$5\cdot {\left(3^{|5-x|}\right)}^2-\frac{50}{3}\cdot 3^{|5-x|}=-5$  

Podstawmy  $3^{|5-x|}=t$  

$5t^2-\frac{50}{3}t=-5\mid +5$  

$5t^2-\frac{50}{3}t+5=0\mid \cdot 3$  

$15t^2-50t+15=0\mid :5$  

$3t^2-10t+3=0$  

$\Delta ={\left(-10\right)}^2-4\cdot 3\cdot 3=100-36=64$  

$t_1=\frac{10-8}{2\cdot 3}=\frac{2}{6}=\frac{1}{3}$  

$t_2=\frac{10+8}{2\cdot 3}=\frac{18}{6}=3$  

Zatem otrzymujemy:

$3^{|5-x|}=\frac{1}{3}\vee 3^{|5-x|}=3$  

$3^{|5-x|}=3^{-1}\vee 3^{|5-x|}=3^1$  

$\left|5-x\right|=-1\vee \left|5-x\right|=1$  

$\text{sprzecznosˊcˊ}\vee 5-x=1\vee 5-x=-1$  

$5-1=x\vee 5+1=x$  

$4=x\vee 6=x$