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Zadatak 0812

Test: MKsIVBGI_073

 

Ako je  \( f(x)=\frac{2x^2-x+1}{2x^4-x^3-x^2+x-1}\), odredi \( f\left ( \sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}-5 \right )\).

 

Rešenje:


 

 

Pojednostavimo koren:

$$\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}= \sqrt{13+30\sqrt{2+\sqrt{(2\sqrt{2})^2+2\cdot 2\sqrt{2}\cdot 1+1^2}}}$$

$$= \sqrt{13+30\sqrt{2+2\sqrt{2}+1}}$$

$$=\sqrt{13+30\sqrt{(\sqrt{2})^2+2\cdot \sqrt{2}\cdot 1+1^2}}$$

$$=\sqrt{13+30(\sqrt{2}+1)}$$

$$=\sqrt{25+2\cdot 5\cdot 3\sqrt{2}+(3\sqrt{2})^2}$$

$$=5+3\sqrt{2}$$

 

$$f\left ( \sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}-5 \right )=f(3\sqrt{2})$$

Pojednostavimo i funkciju:

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

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

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

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

Konačno:

$$f(3\sqrt{2})=\frac{1}{(3\sqrt{2})^2-1}$$

$$=\frac{1}{17}$$


 

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