Thay \(1=\left(x+y\right)^3\) vào biểu thức \(A\) ta có:

\(A=\dfrac{\left(x+y\right)^3}{x^3+y^3}+\dfrac{\left(x+y\right)^3}{xy}\)

\(A=\dfrac{x^3+y^3+3xy.\left(x+y\right)}{x^3+y^3}+\dfrac{x^3+y^3+3xy.\left(x+y\right)}{xy}\)

\(A=1+\dfrac{3xy}{x^3+y^3}+3+\dfrac{x^3+y^3}{xy}\)

\(A=4+\left(\dfrac{3xy}{x^3+y^3}+\dfrac{x^3+y^3}{xy}\right)\ge4+2\sqrt{\dfrac{3xy.\left(x^3+y^3\right)}{xy.\left(x^3+y^3\right)}}\)

\(A=4+2\sqrt{3}\) (Áp dụng BĐT Cauchy cho 2 số dương)

\(\Rightarrow A_{min}=\left(\sqrt{3}+1\right)^2\) khi \(\dfrac{3xy}{x^3+y^3}=\dfrac{x^3+y^3}{xy}\)

\(\Leftrightarrow x^3+y^3=xy\sqrt{3}\)

\(\Leftrightarrow\left(x+y\right).\left(x^2-xy+y^2\right)=xy\sqrt{3}\)

\(\Leftrightarrow x^2+y^2-xy\left(\sqrt{3}+1\right)=0\) và \(x+y=1\).

Đến đây thay \(x=1-y\) vào pt trên ta có:

\(y^2.\left(3+\sqrt{3}\right)-y\left(3+\sqrt{3}\right)+1=0\) có:

\(\Delta=\left(3+\sqrt{3}\right)^2-4.\left(3+\sqrt{3}\right)=2\sqrt{3}\)

\(\Rightarrow\sqrt{\Delta}=\sqrt{2\sqrt{3}}\)

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

\(\Rightarrow x=1-y=\dfrac{3+\sqrt{3}-\sqrt{2\sqrt{3}}}{2.\left(3+\sqrt{3}\right)}\)

Thấy mấy vụ này nhìu r, có nhìu bác lập cả 6,7 acc để tick :v

\(S=\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+...+\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{\left(2n-2\right).2n}\)

\(S=\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{\left(2n-2\right).2n}\right).\dfrac{1}{2}\)

\(S=\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{12}+...+\dfrac{1}{2n-2}-\dfrac{1}{2n}\right).\dfrac{1}{2}\)

\(S=\left(\dfrac{1}{2}-\dfrac{1}{2n}\right).\dfrac{1}{2}\)

\(S=\dfrac{1}{4}-\dfrac{1}{2n.2}< \dfrac{1}{4}\)

\(S=\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+...+\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{4}\left(đpcm\right)\)

\(S=1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)

Mà \(1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}=1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=2-\dfrac{1}{100}< 2\)

\(\Rightarrow\) \(S< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)

Vậy \(S< 2\left(đpcm\right).\)

Sửa thì đem tới sửa chứ có sao đâu mà bạn phải làm quá vậy. Bạn đang spam đó.