Ôn tập chương 1: Căn bậc hai. Căn bậc ba

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Minh Ngọc

Rút gọn a)   \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

b)\(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{961\sqrt{960}+960\sqrt{961}}\)

Hoàng Phú Thiện
2 tháng 9 2022 lúc 14:54

a) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+2\right)+\left(2+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+2}\)

\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+2\right)+\left(\sqrt{2}.\sqrt{2}+\sqrt{3}.\sqrt{2}+2\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+2}\)

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

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

\(=1+\sqrt{2}\)

b) Ta có:

\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}\)

\(=\dfrac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)

\(=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}\)

\(=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}\left(n+1-n\right)}\)

\(=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}\)

\(=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)

Khi đó:

\(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{961\sqrt{960}+960\sqrt{961}}\)

\(=\left(\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}\right)+\left(\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}\right)+...+\left(\dfrac{1}{\sqrt{960}}-\dfrac{1}{\sqrt{961}}\right)\)

\(=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{960}}-\dfrac{1}{\sqrt{961}}\)

\(=1-\dfrac{1}{\sqrt{961}}\)

\(=\dfrac{\sqrt{961}-1}{\sqrt{961}}\)

\(=\dfrac{31-1}{31}\)

\(=\dfrac{30}{31}\)

Trần Tuấn Hoàng
2 tháng 9 2022 lúc 14:56

a) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+2\right)+\left(\sqrt{6}+\sqrt{8}+2\right)}{\sqrt{2}+\sqrt{3}+2}\)

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

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

b) \(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{961\sqrt{960}+960\sqrt{961}}\)

\(=\dfrac{2\sqrt{1}-1\sqrt{2}}{2^2.1-1^2.2}+\dfrac{3\sqrt{2}-2\sqrt{3}}{3^2.2-2^2.3}+...+\dfrac{961\sqrt{960}-960\sqrt{961}}{961^2.960-960^2.961}\)

\(=\dfrac{2\sqrt{1}-1\sqrt{2}}{2.1\left(2-1\right)}+\dfrac{3\sqrt{2}-2\sqrt{3}}{3.2\left(3-2\right)}+...+\dfrac{961\sqrt{960}-960\sqrt{961}}{961.960\left(961-960\right)}\)

\(=\dfrac{2\sqrt{1}-1\sqrt{2}}{2.1}+\dfrac{3\sqrt{2}-2\sqrt{3}}{3.2}+...+\dfrac{961\sqrt{960}-960\sqrt{961}}{961.960}\)

\(=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{960}}-\dfrac{1}{\sqrt{961}}\)

\(=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{961}}=1-\dfrac{1}{31}=\dfrac{30}{31}\)


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