a)

$\sqrt[6]{4^x\cdot {\left(0,125\right)}^{\frac{1}{x}}}=\frac{4\sqrt[3]{2}}{{\left(\sqrt{2}\right)}^x}$  

Zał:

$4^x\cdot {\left(0,125\right)}^{\frac{1}{x}}>0,{\sqrt{2}}^x\ne 0$  

$4^x\cdot {\left(\frac{1}{8}\right)}^{\frac{1}{x}}>0,2^{\frac{1}{2}x}\ne 0$  

$x\ne 0$  

 

$\sqrt[6]{{\left(2^2\right)}^x\cdot {\left(\frac{1}{8}\right)}^{\frac{1}{x}}}=\frac{2^2\cdot 2^{\frac{1}{3}}}{{\left(2^{\frac{1}{2}}\right)}^x}$  

$\sqrt[6]{2^{2x}\cdot {\left(2^{-3}\right)}^{\frac{1}{x}}}=\frac{2^{2\frac{1}{3}}}{2^{\frac{1}{2}x}}$  

$\sqrt[6]{2^{2x-\frac{3}{x}}}=2^{2\frac{1}{3}-\frac{1}{2}x}$  

${\left(2^{2x-\frac{3}{x}}\right)}^{\frac{1}{6}}=2^{\frac{7}{3}-\frac{1}{2}x}$  

$2^{\frac{1}{3}x-\frac{1}{2x}}=2^{\frac{7}{3}-\frac{1}{2}x}$  

$\frac{1}{3}x-\frac{1}{2x}=\frac{7}{3}-\frac{1}{2}x\mid \cdot 6x$  

$2x^2-3=14x-3x^2\mid -14x+3x^2$  

$5x^2-14x-3=0$  

$\Delta ={\left(-14\right)}^2-4\cdot 5\cdot \left(-3\right)=196+60=256$  

$\sqrt{\Delta }=16$  

$x_1=\frac{14-16}{2\cdot 5}=\frac{-2}{10}=-\frac{1}{5}$  

$x_2=\frac{14+16}{2\cdot 5}=\frac{30}{10}=3$  

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b)

$2^{\frac{3x-9}{3x-7}}\cdot \sqrt[3]{{\left(0,25\right)}^{\frac{3x-1}{2x-2}}}=1$  

Zał:

${\left(0,25\right)}^{\frac{3x-1}{2x-2}}>0$  

${\left(\frac{1}{4}\right)}^{\frac{3x-1}{2x-2}}>0$  

$2x-2\ne 0$  

$x\ne 1$  

$3x-7\ne 0$  

$3x\ne 7$  

$x\ne \frac{7}{3}$  

 

$2^{\frac{3x-9}{3x-7}}\cdot \sqrt[3]{{\left(\frac{1}{4}\right)}^{\frac{3x-1}{2x-2}}}=2^0$  

$2^{\frac{3x-9}{3x-7}}\cdot \sqrt[3]{{\left(2^{-2}\right)}^{\frac{3x-1}{2x-2}}}=2^0$  

$2^{\frac{3x-9}{3x-7}\cdot \sqrt[3]{2^{\frac{-3x+1}{x-1}}}=2^0}$  

$2^{\frac{3x-9}{3x-7}}\cdot {\left(2^{\frac{-3x+1}{x-1}}\right)}^{\frac{1}{3}}=2^0$  

$2^{\frac{3x-9}{3x-7}}\cdot 2^{\frac{-3x+1}{3x-3}}=2^0$  

$2^{\frac{3x-9}{3x-7}+\frac{-3x+1}{3x-3}}=2^0$  

$\frac{3x-9}{3x-7}+\frac{-3x+1}{3x-3}=0\mid \cdot \left(3x-7\right)\left(3x-3\right)$  

$\left(3x-9\right)\left(3x-3\right)+\left(-3x+1\right)\left(3x-7\right)=0$  

$9x^2-9x-27x+27-9x^2+21x+3x-7=0$  

$-12x+20=0\mid -20$  

$-12x=-20\mid :\left(-12\right)$  

$x=\frac{20}{12}$  

$x=\frac{5}{3}$  

$x=1\frac{2}{3}$  

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c)

Zał:

${\left(0,25\right)}^{5-\frac{x}{4}}\ge 0\wedge x+1\ge 0$  

${\left(\frac{1}{4}\right)}^{5-\frac{x}{4}}\ge 0\wedge x\ge -1$  

${\left(2^{-2}\right)}^{5-\frac{x}{4}}\ge 0\wedge x\ge -1$  

$2^{-10+\frac{1}{2}x}\ge 0\wedge x\ge -1$  

$2^{-10}\cdot 2^{\frac{1}{2}x}\ge 0\wedge x\ge -1$  

$2^{\frac{1}{2}x}\ge 0\wedge x\ge -1$  

${\sqrt{2}}^x\ge 0\wedge x\ge -1$  

$x\in \mathbb{R}\wedge x\ge -1$  

 

$\sqrt{{\left(0,25\right)}^{5-\frac{x}{4}}}=2^{\sqrt{x+1}-4}$  

$\sqrt{{\left(\frac{1}{4}\right)}^{5-\frac{x}{4}}}=2^{\sqrt{x+1}-4}$  

$\sqrt{{\left(2^{-2}\right)}^{5-\frac{x}{4}}}=2^{\sqrt{x+1}-4}$  

$\sqrt{2^{-10+\frac{1}{2}x}}=2^{\sqrt{x+1}-4}$  

${\left(2^{-10+\frac{1}{2}x}\right)}^{\frac{1}{2}}=2^{\sqrt{x+1}-4}$  

$2^{-5+\frac{1}{4}x}=2^{\sqrt{x+1}-4}$  

$-5+\frac{1}{4}x=\sqrt{x+1}-4\mid +4$  

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

Uwaga! Zauważmy, że z prawej strony mamy liczbę nieujemną, zatem z lewej strony również musi być liczba nieujemna, otrzymujemy:

$-1+\frac{1}{4}x\ge 0\mid \cdot 4$  

$-4+x\ge 0\mid +4$  

$x\ge 4$  

Wracając do równania otrzymujemy:

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

$\frac{1}{16}{x}^2-\frac{1}{2}x+1=x+1\mid \cdot 16$  

$x^2-8x+16=16x+16\mid -16x-16$  

$x^2-24x=0$  

$x\left(x-24\right)=0$  

$x=0\vee x-24=0$  

$\text{sprzecznosˊcˊ}\vee x=24$  

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d)

Zał:

$9-x>0$  

$9>x$  

 

${\left(\frac{5}{2}\right)}^{\sqrt{9-x}-1}={\left(0,4\right)}^{\frac{4+\sqrt{9-x}}{\sqrt{9-x}}-5}$  

${\left({\left(\frac{2}{5}\right)}^{-1}\right)}^{\sqrt{9-x}-1}={\left(\frac{4}{10}\right)}^{\frac{4+\sqrt{9-x}}{\sqrt{9-x}}-5}$  

${\left(\frac{2}{5}\right)}^{-\sqrt{9-x}+1}={\left(\frac{2}{5}\right)}^{\frac{4+\sqrt{9-x}}{\sqrt{9-x}}-5}$  

$-\sqrt{9-x}+1=\frac{4+\sqrt{9-x}}{\sqrt{9-x}}-5\mid \cdot \sqrt{9-x}$  

$-{\sqrt{9-x}}^2+\sqrt{9-x}=4+\sqrt{9-x}-5\sqrt{9-x}\mid +{\sqrt{9-x}}^2-\sqrt{9-x}$  

$0={\sqrt{9-x}}^2-5\sqrt{9-x}+4$  

Podstawiając  $\sqrt{9-x}=t,t>0$  otrzymujemy:

$0=t^2-5t+4$  

$\Delta ={\left(-5\right)}^2-4\cdot 1\cdot 4=25-16=9$  

$t_1=\frac{5-3}{2}=\frac{2}{2}=1$  

$t_2=\frac{5+3}{2}=\frac{8}{2}=4$  

$\sqrt{9-x}=1\vee \sqrt{9-x}=4$  

$9-x=1\vee 9-x=16$  

$-x=-8\vee -x=7$  

$x=8\vee x=-7$