[tex]x=2\sqrt{2}-1\ \ \ \ i\ \ \ \ y=\sqrt{2}+1\\\\\\x+y=2\sqrt{2}-1+\sqrt{2}+1=3\sqrt{2}\\\\\\x-y=2\sqrt{2}-1-(\sqrt{2}+1)=2\sqrt{2}-1-\sqrt{2}-1=\sqrt{2}-2\\\\\\x\cdot y=(2\sqrt{2}-1)(\sqrt{2}+1)=2\sqrt{2}\cdot\sqrt{2}+2\sqrt{2}\cdot1-1\cdot\sqrt{2}-1\cdot1=\\\\=2\cdot2+2\sqrt{2}-\sqrt{2}-1=4+\sqrt{2}-1=3+\sqrt{2}[/tex]
[tex]\dfrac{x}{y}=\dfrac{2\sqrt{2}-1}{\sqrt{2}+1}=\dfrac{2\sqrt{2}-1}{\sqrt{2}+1}\ \cdot\dfrac{\sqrt{2}-1}{\sqrt{2}-1}=\dfrac{(2\sqrt{2}-1)(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)}=\dfrac{2\sqrt{2}\cdot\sqrt{2}-2\sqrt{2}-\sqrt{2}+1}{(\sqrt{2})^2-1^2}=\\\\\\=\dfrac{2\cdot2-3\sqrt{2}+1}{2-1}=\dfrac{4-3\sqrt{2}+1}{1}=\dfrac{5-3\sqrt{2}}{1}=5-3\sqrt{2}[/tex]
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[tex]x=2\sqrt{2}-1\ \ \ \ i\ \ \ \ y=\sqrt{2}+1\\\\\\x+y=2\sqrt{2}-1+\sqrt{2}+1=3\sqrt{2}\\\\\\x-y=2\sqrt{2}-1-(\sqrt{2}+1)=2\sqrt{2}-1-\sqrt{2}-1=\sqrt{2}-2\\\\\\x\cdot y=(2\sqrt{2}-1)(\sqrt{2}+1)=2\sqrt{2}\cdot\sqrt{2}+2\sqrt{2}\cdot1-1\cdot\sqrt{2}-1\cdot1=\\\\=2\cdot2+2\sqrt{2}-\sqrt{2}-1=4+\sqrt{2}-1=3+\sqrt{2}[/tex]
[tex]\dfrac{x}{y}=\dfrac{2\sqrt{2}-1}{\sqrt{2}+1}=\dfrac{2\sqrt{2}-1}{\sqrt{2}+1}\ \cdot\dfrac{\sqrt{2}-1}{\sqrt{2}-1}=\dfrac{(2\sqrt{2}-1)(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)}=\dfrac{2\sqrt{2}\cdot\sqrt{2}-2\sqrt{2}-\sqrt{2}+1}{(\sqrt{2})^2-1^2}=\\\\\\=\dfrac{2\cdot2-3\sqrt{2}+1}{2-1}=\dfrac{4-3\sqrt{2}+1}{1}=\dfrac{5-3\sqrt{2}}{1}=5-3\sqrt{2}[/tex]