Question

[Phan Nghĩa]

Cho 2 số x,y thỏa mãn \(x^3-3xy^2=10\)và \(y^3-3yx^2=30\)

Tính \(p=x^2+y^2\)

Answer 1

[Nguyễn Minh Đăng]

Ta có: \(\hept{\begin{cases}x^3-3xy^2=10\\y^3-3x^2y=30\end{cases}}\Leftrightarrow\hept{\begin{cases}\left(x^3-3xy^2\right)^2=100\\\left(y^3-3x^2y\right)^2=900\end{cases}}\)

\(\Rightarrow\left(x^3-3xy^2\right)^2+\left(y^3-3x^2y\right)^2=1000\)

\(\Leftrightarrow x^6-6x^4y^2+9x^2y^4+y^6-6x^2y^4+9x^4y^2=1000\)

\(\Leftrightarrow x^6+3x^4y^2+3x^2y^4+y^6=1000\)

\(\Leftrightarrow\left(x^2+y^2\right)^3=1000\)

\(\Rightarrow x^2+y^2=10\)

Answer 2

[Fucking Out]

Có: \(x^3-3xy^2=10\)

=> \(x^6+9x^2y^4-6x^4y^2=100\left(1\right)\)

Có: \(y^3-3yx^2=30\)

=> \(y^6-6y^4x^2+9x^4y^2=900\left(2\right)\)

Lấy (1) + (2) ta được:

=> \(x^6+y^6+3x^2y^4+3x^4y^2=1000\)

=> \(\left(x^2+y^2\right)^3=1000\)

=> \(x^2+y^2=10\)

=> \(p=10.\)