\(x=\sqrt{2+\sqrt{\frac{5+\sqrt{5}}{2}}}+\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}\)

\(\Leftrightarrow x^2=4+2\sqrt{\frac{3-\sqrt{5}}{2}}\)

\(\Leftrightarrow x^2=4+\sqrt{6-2\sqrt{5}}\)

\(\Leftrightarrow x^2=4+\sqrt{5}-1\)

\(\Leftrightarrow x^2=3+\sqrt{5}\)

\(\Leftrightarrow x^2=\frac{6+2\sqrt{5}}{2}\)

\(\Leftrightarrow x=\frac{\sqrt{5}+1}{\sqrt{2}}\)

Ta có:

\(P=\left(x-\sqrt{3-\sqrt{5}}-\sqrt{2}+1\right)^{2019}\)

\(=\left(\frac{\sqrt{5}+1}{\sqrt{2}}-\sqrt{3-\sqrt{5}}-\sqrt{2}+1\right)^{2019}\)

\(=\left(\frac{\sqrt{5}+1-\sqrt{6-2\sqrt{5}}-2+\sqrt{2}}{\sqrt{2}}\right)^{2019}\)

\(=\left(\frac{\sqrt{5}+1-\left(\sqrt{5}-1\right)-2+\sqrt{2}}{\sqrt{2}}\right)^{2019}\)

\(=\left(1\right)^{2019}=1\)   

Ta có x³ < y³ (1)

Lại có (x + 2)³ = x³ + 6x² + 12x + 8 = (x³ + x² + x + 1) + (5x² + 11x + 7) 

=  \(y^3+5\left(x+\frac{11}{10}\right)^2+\frac{19}{20}\)

Nhận thấy \(5\left(x+\frac{11}{10}\right)^2+\frac{19}{20}>0\)

=> \(y^3+5\left(x+\frac{11}{10}\right)^3+\frac{19}{20}>y^3\)

=> \(\left(x+2\right)^3>y^3\)(2)

Từ (1) và (2) => y = (x + 1)³ 

Khi đó 1 + x + x² + x³ = x³ + 3x² + 3x + 1 

<=> 2x² + 2x = 0

<=> 2x(x + 1) = 0

<=> \(\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

Khi x = 0 => y = 1

Khi x = -1 => y = 0

Vậy các cặp (x ; y) thỏa mãn là (0;1) ; (-1;0) 

\(A=\frac{9\left(a+b+c\right)^2}{ab+bc+ca}+\frac{a^2+b^2+c^2}{ab+bc+ca}\)

\(A=\frac{9\left(a^2+b^2+c^2+2ab+2bc+2ca\right)}{ab+bc+ca}+\frac{a^2+b^2+c^2}{ab+bc+ca}\)

\(A=\frac{9a^2+9b^2+9c^2+18ab+18bc+18ca}{ab+bc+ca}+\frac{a^2+b^2+c^2}{ab+bc+ca}\)

\(A=\frac{9a^2+9b^2+9c^2+18ab+18bc+18ca+a^2+b^2+c^2}{ab+bc+ca}\)

dễ thấy \(9a^2+9b^2+9c^2\ge9ab+9bc+9ca\)(bđt tương đương)

\(a^2+b^2+c^2\ge ab+bc+ca\)

\(A\ge\frac{28ab+28bc+28ca}{ab+bc+ca}=28\)dấu "=" xảy ra khi và chỉ khi \(a=b=c=1\)

\(< =>MIN:A=28\)

Đặt \(ax^3=by^3=cz^3=k\).

\(\sqrt[3]{ax^2+by^2+cz^2}=\sqrt[3]{\frac{ax^3}{x}+\frac{by^3}{y}+\frac{cz^3}{z}}=\sqrt[3]{k}\)

\(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=\frac{\sqrt[3]{ax^3}}{x}+\frac{\sqrt[3]{by^3}}{y}+\frac{\sqrt[3]{cz^3}}{z}=\sqrt[3]{k}\)

Do đó ta có đpcm.