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

\(\Rightarrow A^2=\left(2-\sqrt{3}\right)^2\left(2+\sqrt{3}\right)+2\left(2-\sqrt{3}\right)\sqrt{2+\sqrt{3}}\left(2+\sqrt{3}\right)\sqrt{2-\sqrt{3}}+\left(2+\sqrt{3}\right)^2\left(2-\sqrt{3}\right)\)

\(=\left(2-\sqrt{3}\right)+2.1.1+\left(2+\sqrt{3}\right)\)

\(=2-\sqrt{3}+2+2+\sqrt{3}\)

\(=6\)

\(\Rightarrow A=\sqrt{6}\)

a, \(x^2+6x-4y^2+9=\left(x+3\right)^2-4y^2\)

\(=\left(x-2y+3\right)\left(x+2y+3\right)\)

b, \(x^2-7x+xy-7y=x\left(x-7\right)+y\left(x-7\right)=\left(x+y\right)\left(x-7\right)\)

Ta có: \(x^2-xy+y^2=x^2-2.xy.\dfrac{1}{2}+\dfrac{1}{4}y^2+\dfrac{3}{4}y^2\)

\(\left(x-\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2\ge0\forall x,y\)

\(\Rightarrowđpcm\)

Ta có: \(\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{2017^3}< \dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{2015.2016.2017}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{2015.2016.2017}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{2015.2016}-\dfrac{1}{2016.2017}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2016.2017}\right)\)

\(=\dfrac{1}{4}-\dfrac{1}{2.2016.2017}< \dfrac{1}{4}\)

\(\Rightarrowđpcm\)

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