Đặt \(a=2^{\frac{1}{3}}\)\(\Rightarrow\hept{\begin{cases}x=\frac{a}{a^2+a+1}\\y=\frac{a}{a^2-a+1}\end{cases}}\)

\(A{=xy(y^2-x^2)\\=xy(y+x)(y-x)\\=\dfrac{a^2}{a^4+a^2+1}\dfrac{2a^3+2a}{a^4+a^2+1}\dfrac{2a^2}{a^4+a^2+1}\\=\dfrac{8a^2(a^2+1)}{(a+1)^6}\\=\dfrac{8a^2(a^2+1)}{(a^3+3a^2+3a+1)^2}\\=\dfrac{8a^2(a^2+1)}{9(a^2+a+1)^2}}\)

Vì \(\left(a-1\right)\left(a^2+a+1\right)=a^3-1=1\). khi đó 

\(A=\dfrac{8}{9}a^2(a^2+1)(a-1)^2=\dfrac{8}{9}a^2(a^4-2a^3+a^2+a^2-2a+1)=\dfrac{8}{9}a^2(2a^2-3)=\dfrac{8}{9}(4a-3a^2)\)

Câu hỏi của bạn thiếu rất nhiều giả thiết. Tam giác ABC có tâm O, ý của bạn là trực tâm hay trọng tâm . Rồi quan trọng hơn. chứng minh KE vuông với DE, nhưng không cho điểm D thì rồi chứng minh kiểu gì đây hả bạn? :/

Ta có : x - y = 2 => x=2+y (1)

 Mà 5x-3y=10 (2)

Thay (1) vào (2) ta dc : 5(2+y) - 3y =10

                                 => y = 0

                                 => x =0+2=2

\(5x-3y=10\)

\(\Leftrightarrow3\left(x-y\right)+2x=10\)

\(\Leftrightarrow6+2x=10\)

\(\Leftrightarrow x=2\)

a) \(\frac{1}{2+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}\)

\(=\frac{3\sqrt{3}+3}{\sqrt{3}.\left(2+\sqrt{3}\right)\left(3+\sqrt{3}\right)}+\frac{9+5\sqrt{3}}{\sqrt{3}.\left(2+\sqrt{3}\right)\left(3+\sqrt{3}\right)}-\frac{2.\left(2\sqrt{3}+3\right)}{\sqrt{3}.\left(2+\sqrt{3}\right)\left(3+\sqrt{3}\right)}\)

\(=\frac{3\sqrt{3}+3+9+5\sqrt{3}-2\left(2\sqrt{3}+3\right)}{\sqrt{3}.\left(2+\sqrt{3}\right)\left(3+\sqrt{3}\right)}\)

\(=\frac{-2\left(2\sqrt{3}+3\right)+8\sqrt{3}+3+9}{\sqrt{3}.\left(2+\sqrt{3}\right)\left(3+\sqrt{3}\right)}\)

\(=\frac{-2\left(2\sqrt{3}+3\right)+8\sqrt{3}+3+9}{9\sqrt{3}+15}\)

\(=\frac{4\sqrt{3}+6}{9\sqrt{3}+15}\)

\(=\frac{3-\sqrt{3}}{3}\)

b) \(\frac{\sqrt{5}-2}{5+2\sqrt{5}}-\frac{1}{2+\sqrt{5}}+\frac{1}{\sqrt{5}}\)

\(=\frac{\left(\sqrt{5}-2\right)\left(2\sqrt{5}+5\right)}{\sqrt{5}.\left(2+\sqrt{5}\right)\left(5+2\sqrt{5}\right)}-\frac{5\sqrt{5}+10}{\sqrt{5}.\left(2+\sqrt{5}\right)\left(5+2\sqrt{5}\right)}+\frac{20+9\sqrt{5}}{\sqrt{5}.\left(2+\sqrt{5}\right)\left(5+2\sqrt{5}\right)}\)

\(=\frac{\left(\sqrt{5}-2\right)\left(2\sqrt{5}+5\right)-\left(5\sqrt{5}+10\right)+20-9\sqrt{5}}{20\sqrt{5}+45}\)

\(=\frac{5\sqrt{5}+10}{20\sqrt{5}+45}\)

\(=\frac{5\left(\sqrt{5}+2\right)}{5\left(4\sqrt{5}+9\right)}\)

\(=\frac{\sqrt{5}+2}{4\sqrt{5}+9}\)

\(=\sqrt{5}-2\)