Question

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Answer 1

 

$\displaystyle\bf\\Se~dau~numerele:\\\\x=\sqrt{4-\sqrt{7}}\\\\y=\sqrt{4+\sqrt{7}}\\\\Se~cere:\\\\a)~~x\cdot y=?\\\\ b)~~(x-y)^2=?\\\\Rezolvare:\\\\a)\\\\x\cdot y=\sqrt{4-\sqrt{7}}\cdot \sqrt{4+\sqrt{7}}\\\\x\cdot y=\sqrt{\bigg(4-\sqrt{7}\bigg)\cdot \bigg(4+\sqrt{7}\bigg)} \\\\\\x\cdot y=\sqrt{4^2-\bigg(\sqrt{7}\bigg)^2}\\\\\\x\cdot y=\sqrt{16-7}\\\\x\cdot y=\sqrt{9}\\\\\boxed{\bf x\cdot y= 3}$

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$\displaystyle\bf\\b)\\\\(x-y)^2=\left(\sqrt{4-\sqrt{7}}- \sqrt{4+\sqrt{7}}\right)^2\\\\\\(x-y)^2=\left(\sqrt{4-\sqrt{7}}\right)^2-2\left(\sqrt{4-\sqrt{7}}\cdot \sqrt{4+\sqrt{7}}\right) +\left(\sqrt{4+\sqrt{7}}}~\right)^2\\\\\\(x-y)^2=\left(\sqrt{4-\sqrt{7}}\right)^2-2\cdot 3 +\left(\sqrt{4+\sqrt{7}}}~\right)^2\\\\(x-y)^2=(4-\sqrt{7})-2\cdot 3 +(4+\sqrt{7})\\\\(x-y)^2=4+4 -\sqrt{7}+\sqrt{7}-6\\\\(x-y)^2=8-6\\\\\boxed{\bf (x-y)^2=2}$