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Question

rút gọn $\dfrac{1}{\sqrt{2}+1}$ - $\dfrac{\sqrt{8}-\sqrt{10}}{2-\sqrt{5}}$

YangSu · Accepted Answer

$\dfrac{1}{\sqrt{2}+1}-\dfrac{\sqrt{8}-\sqrt{10}}{2-\sqrt{5}}\
=\dfrac{1}{\sqrt{2}+1}-\dfrac{\sqrt{2}.\sqrt{4}-\sqrt{2}.\sqrt{5}}{2-\sqrt{5}}\
=\dfrac{1}{\sqrt{2}+1}-\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{2-\sqrt{5}}\
=\dfrac{1}{\sqrt{2}+1}-\sqrt{2}\
=\dfrac{1-\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\
=\dfrac{1-2-\sqrt{2}}{\sqrt{2}+1}\
=\dfrac{-\sqrt{2}-1}{\sqrt{2}+1}\
=\dfrac{-\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\
=-1$Câu trả lời: $\dfrac{1}{\sqrt{2}+1}-\dfrac{\sqrt{8}-\sqrt{10}}{2-\sqrt{5}}\
=\dfrac{1}{\sqrt{2}+1}-\dfrac{\sqrt{2}.\sqrt{4}-\sqrt{2}.\sqrt{5}}{2-\sqrt{5}}\
=\dfrac{1}{\sqrt{2}+1}-\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{2-\sqrt{5}}\
=\dfrac{1}{\sqrt{2}+1}-\sqrt{2}\
=\dfrac{1-\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\
=\dfrac{1-2-\sqrt{2}}{\sqrt{2}+1}\
=\dfrac{-\sqrt{2}-1}{\sqrt{2}+1}\
=\dfrac{-\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\
=-1$

T . Anhh · Answer

$\dfrac{1}{\sqrt{2}+1}-\dfrac{\sqrt{8}-\sqrt{10}}{2-\sqrt{5}}=-1+\sqrt{2}-\sqrt{2}=-1$Câu trả lời: $\dfrac{1}{\sqrt{2}+1}-\dfrac{\sqrt{8}-\sqrt{10}}{2-\sqrt{5}}=-1+\sqrt{2}-\sqrt{2}=-1$

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