Violympic toán 9

Không Biết Chán

giải B=\(\left(\dfrac{1}{\sqrt{3}-\sqrt{2}}-\dfrac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)\)

Nguyễn Linh
13 tháng 10 2018 lúc 21:43

B = \(\left(\dfrac{1}{\sqrt{3}-\sqrt{2}}-\dfrac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)\)

= \(\left[\dfrac{1-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(1-\sqrt{2}\right)}-\dfrac{\left(\sqrt{3}-\sqrt{6}\right)\left(\sqrt{3}-\sqrt{2}\right)}{\left(\sqrt{3}-\sqrt{2}\right)\left(1-\sqrt{2}\right)}-\sqrt{3}\right]\) \(\left(\sqrt{2}+\sqrt{3}\right)\)

= \(\left[\dfrac{1-\sqrt{2}-\left(3-\sqrt{6}-3\sqrt{2}+2\sqrt{3}\right)}{\left(\sqrt{3}-\sqrt{2}\right)\left(1-\sqrt{2}\right)}-\sqrt{3}\right]\) \(\left(\sqrt{2}+\sqrt{3}\right)\)

= \(\left[\dfrac{1-\sqrt{2}-3+\sqrt{6}+3\sqrt{2}-2\sqrt{3}}{\left(\sqrt{3}-\sqrt{2}\right)\left(1-\sqrt{2}\right)}-\sqrt{3}\right]\)\(\left(\sqrt{2}+\sqrt{3}\right)\)

= \(\left[\dfrac{-2+2\sqrt{2}+\sqrt{6}-2\sqrt{3}}{\left(\sqrt{3}-\sqrt{2}\right)\left(1-\sqrt{2}\right)}-\sqrt{3}\right]\)\(\left(\sqrt{2}+\sqrt{3}\right)\)

= \(\left[\dfrac{-2\left(1-\sqrt{2}\right)+\sqrt{6}\left(1-\sqrt{2}\right)}{\left(\sqrt{3}-\sqrt{2}\right)\left(1-\sqrt{2}\right)}-\sqrt{3}\right]\)\(\left(\sqrt{2}+\sqrt{3}\right)\)

= \(\left[\dfrac{\left(-2+\sqrt{6}\right)\left(1-\sqrt{2}\right)}{\left(\sqrt{3}-\sqrt{2}\right)\left(1-\sqrt{2}\right)}-\sqrt{3}\right]\)\(\left(\sqrt{2}+\sqrt{3}\right)\)

= \(\left[\dfrac{-\sqrt{4}+\sqrt{6}}{\sqrt{3}-\sqrt{2}}-\sqrt{3}\right]\)\(\left(\sqrt{2}+\sqrt{3}\right)\)

= \(\left[\dfrac{-\sqrt{2}\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{3}-\sqrt{2}}-\sqrt{3}\right]\)\(\left(\sqrt{2}+\sqrt{3}\right)\)

= (\(\sqrt{2}-\sqrt{3}\))\(\left(\sqrt{2}+\sqrt{3}\right)\) = -1

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tran nguyen bao quan
14 tháng 10 2018 lúc 10:26

B=\(\left(\dfrac{1}{\sqrt{3}-\sqrt{2}}-\dfrac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)=\left(\dfrac{\sqrt{3}+\sqrt{2}}{3-2}-\dfrac{\sqrt{3}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)=\left(\sqrt{3}+\sqrt{2}-\sqrt{3}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)=\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)=2-3=-1\)

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