√6 + √2/√3 +1 + 4/√2 - 3/√2-1

1 câu trả lời

$\dfrac{\sqrt[]{6}+\sqrt[]{2}}{\sqrt[]{3}+1}+\dfrac{4}{\sqrt[]{2}}-\dfrac{3}{\sqrt[]{2}-1}$ 
$=\dfrac{\sqrt[]{2}(\sqrt[]{3}+1)}{\sqrt[]{3}+1}+\dfrac{4\sqrt[]{2}}{\sqrt[]{2}^2}-\dfrac{3(\sqrt[]{2}+1)}{\sqrt[]{2}^2-1^2}$
$=\sqrt[]{2}+2\sqrt[]{2}-3\sqrt[]{2}-3$
$=-3$

 

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