Odpowiedź :
[tex]\dfrac{2\sqrt{32}-\sqrt{72}}{2\sqrt2}=\dfrac{2\sqrt{16\cdot2}-\sqrt{36\cdot2}}{2\sqrt2}=\dfrac{2\sqrt{4^2\cdot2}-\sqrt{6^2\cdot2}}{2\sqrt2}=\\\\\\=\dfrac{2\cdot4\sqrt2-6\sqrt2}{2\sqrt2}=\dfrac{8\sqrt2-6\sqrt2}{2\sqrt2}=\dfrac{2\sqrt2}{2\sqrt2}=\huge\boxed1[/tex]
Odpowiedź:
[tex]\frac{2\sqrt{32} - \sqrt{72}}{2\sqrt{2}} = \frac{2\sqrt{2^5} - \sqrt{2^3 \cdot 3^2}}{2\sqrt{2}} = \frac{2\cdot 2^2\sqrt{2}- 2\cdot 3\sqrt{2}}{2\sqrt{2}} = \frac{2\cdot 4\sqrt{2} - 6\sqrt{2}}{2\sqrt{2}} = \frac{8\sqrt{2} - 6\sqrt{2}}{2\sqrt{2}} = \frac{2\sqrt{2}}{2\sqrt{2}} = 1[/tex]