Odpowiedź:
[tex]\frac{4}{\sqrt{2}-1 } +\frac{2}{\sqrt{2}+2 }=\frac{4(\sqrt{2}+2)+2(\sqrt{2}-1) }{(\sqrt{2}-1)(\sqrt{2}+2) } =\frac{4\sqrt{2}+8+2\sqrt{2}-2 }{2+2\sqrt{2}-\sqrt{2}-2 }=\frac{6\sqrt{2}+6}{\sqrt{2} }=\frac{6\sqrt{2}+6 }{\sqrt{2} } \cdot \frac{\sqrt{2} }{\sqrt{2} } =\frac{12+6\sqrt{2} }{2} =6+3\sqrt{2}[/tex]
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Odpowiedź:
[tex]\frac{4}{\sqrt{2}-1 } +\frac{2}{\sqrt{2}+2 }=\frac{4(\sqrt{2}+2)+2(\sqrt{2}-1) }{(\sqrt{2}-1)(\sqrt{2}+2) } =\frac{4\sqrt{2}+8+2\sqrt{2}-2 }{2+2\sqrt{2}-\sqrt{2}-2 }=\frac{6\sqrt{2}+6}{\sqrt{2} }=\frac{6\sqrt{2}+6 }{\sqrt{2} } \cdot \frac{\sqrt{2} }{\sqrt{2} } =\frac{12+6\sqrt{2} }{2} =6+3\sqrt{2}[/tex]