Rozwiązujemy nierówność 1.

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

$2\left(\frac{1}{4}{x}^2-x+1\right)\le \frac{1}{2}{x}^2+3\sqrt{2}x+9-3\sqrt{2}x$  

$\frac{1}{2}{x}^2-2x+2\le \frac{1}{2}{x}^2+9\mid -\frac{1}{2}{x}^2$  

$-2x+2\le 9\mid -2$  

$-2x\le 7\mid \cdot \left(-1\right)$  

$2x\ge -7\mid :2$  

$x\ge -\frac{7}{2}$  

Czyli:

$x\in ⟨-3\frac{1}{2},\infty )$  

$x\in ⟨-\frac{7}{2},\infty )$  

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Rozwiązujemy nierówność 2.

${\left(\sqrt{2}x-\sqrt{3}\right)}^2-{\left(\sqrt{6}-x\right)}^2>\left(x-2\sqrt{3}\right)\left(x+3\sqrt{3}\right)$  

$2x^2-2\sqrt{6}x+3-\left(6-2\sqrt{6}+x^2\right)>x^2+3\sqrt{3}x-2\sqrt{3}x-18$  

$2x^2-2\sqrt{6}x+3-6+2\sqrt{6}-x^2>x^2+\sqrt{3}x-18$  

$x^2-3>x^2+\sqrt{3}x-18\mid -x^2$  

$-3>\sqrt{3}x-18\mid +18$  

$15>\sqrt{3}x\mid :\sqrt{3}$  

$\frac{15}{\sqrt{3}}>x$   

Usuwamy niewymierność z mianownika.

$x=\frac{15}{\sqrt{3}}=\frac{15}{\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}}=\frac{15\sqrt{3}}{3}=5\sqrt{3}$  

Czyli:

$x\in \left(-\infty ,5\sqrt{3}\right)$  

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Zatem:

$x\in ⟨-3\frac{1}{2},5\sqrt{3})$  

Zauważmy, że:

$5\sqrt{3}\approx 8,7$  

Czyli:

$a=8$  

$b=-3$  

Obliczamy wartość podanego wyrażenia.

${\log }_{\sqrt{b+5}}{\left(\frac{b+1}{a}\right)}^2={\log }_{\sqrt{-3+5}}{\left(\frac{-3+1}{8}\right)}^2={\log }_{\sqrt{2}}{\left(-\frac{2}{8}\right)}^2={\log }_{\sqrt{2}}{\left(-\frac{1}{4}\right)}^2=$  

$={\log }_{\sqrt{2}}{\left(\frac{1}{4}\right)}^2={\log }_{\sqrt{2}}{\left(\frac{1}{2^2}\right)}^2={\log }_{\sqrt{2}}{\left({\left(\frac{1}{2}\right)}^2\right)}^2={\log }_{\sqrt{2}}({\left(2^{-2}\right)}^2={\log }_{\sqrt{2}}{2}^{\left(-2\right)\cdot 2}=$  

$={\log }_{\sqrt{2}}{2}^{-4}={\log }_{\sqrt{2}}{\left(2^{\frac{1}{2}}\cdot 2\right)}^{-4}={\log }_{\sqrt{2}}{\left(2^{\frac{1}{2}}\right)}^{2\cdot \left(-4\right)}={\log }_{\sqrt{2}}{\left(\sqrt{2}\right)}^{-8}=-8$