Это нужно доказать: ![\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=1;\ x=\sqrt[3]{2+\sqrt{5}};\ y=\sqrt[3]{2-\sqrt{5}};](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B2%2B%5Csqrt%7B5%7D%7D%2B%5Csqrt%5B3%5D%7B2-%5Csqrt%7B5%7D%7D%3D1%3B%5C%20x%3D%5Csqrt%5B3%5D%7B2%2B%5Csqrt%7B5%7D%7D%3B%5C%20y%3D%5Csqrt%5B3%5D%7B2-%5Csqrt%7B5%7D%7D%3B "\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=1;\ x=\sqrt[3]{2+\sqrt{5}};\ y=\sqrt[3]{2-\sqrt{5}};")
x+y=1; - для упрощение доказываем это.
![x^3+y^3=\sqrt[3]{2+\sqrt{5}}^3+\sqrt[3]{2-\sqrt{5}}^3=2+\sqrt{5}+2-\sqrt{5}=4;](https://tex.z-dn.net/?f=x%5E3%2By%5E3%3D%5Csqrt%5B3%5D%7B2%2B%5Csqrt%7B5%7D%7D%5E3%2B%5Csqrt%5B3%5D%7B2-%5Csqrt%7B5%7D%7D%5E3%3D2%2B%5Csqrt%7B5%7D%2B2-%5Csqrt%7B5%7D%3D4%3B "x^3+y^3=\sqrt[3]{2+\sqrt{5}}^3+\sqrt[3]{2-\sqrt{5}}^3=2+\sqrt{5}+2-\sqrt{5}=4;")
x³+y³=(x+y)(x²-xy+y²)=(x+y)(x²+2xy+y²-3xy)=(x+y)((x+y)²-3xy)=4;
![xy=\sqrt[3]{2+\sqrt{5}}*\sqrt[3]{2-\sqrt{5}}=\sqrt[3]{2^2-\sqrt{5}^2}=\sqrt[3]{-1}=-1;](https://tex.z-dn.net/?f=xy%3D%5Csqrt%5B3%5D%7B2%2B%5Csqrt%7B5%7D%7D%2A%5Csqrt%5B3%5D%7B2-%5Csqrt%7B5%7D%7D%3D%5Csqrt%5B3%5D%7B2%5E2-%5Csqrt%7B5%7D%5E2%7D%3D%5Csqrt%5B3%5D%7B-1%7D%3D-1%3B "xy=\sqrt[3]{2+\sqrt{5}}*\sqrt[3]{2-\sqrt{5}}=\sqrt[3]{2^2-\sqrt{5}^2}=\sqrt[3]{-1}=-1;")
x³+y³=(x+y)((x+y)²-3(-1))=(x+y)((x+y)²+3)=4;
(x+y)³+3(x+y)-4=0; Пусть z=x+y;
z³+3z-4=0; найдем корни.
z³-z²+z²-z+4z-4=0;
z²(z-1)+z(z-1)+4(z-1)=0;
(z-1)(z²+z+4)=0; z=1; первый корень уравнения.
z²+z+4=0; D=1-4*4=-15<0; больше действительных корней нет.</p>
![z=1=x+y=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}};\\\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=1;](https://tex.z-dn.net/?f=z%3D1%3Dx%2By%3D%5Csqrt%5B3%5D%7B2%2B%5Csqrt%7B5%7D%7D%2B%5Csqrt%5B3%5D%7B2-%5Csqrt%7B5%7D%7D%3B%5C%5C%5Csqrt%5B3%5D%7B2%2B%5Csqrt%7B5%7D%7D%2B%5Csqrt%5B3%5D%7B2-%5Csqrt%7B5%7D%7D%3D1%3B "z=1=x+y=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}};\\\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=1;")
что и требовалось доказать.