Question

Giúp mik với

Tìm x,y biết \(\dfrac{x^3+y^3}{6}=\dfrac{x^3-2y^3}{4}\) và \(x^6.y^6=64\)

Answer 1

\(\dfrac{x^3+y^3}{6}=\dfrac{x^3-2y^3}{4}\\ \Rightarrow4x^3+4y^3=6x^3-12y^3\\ \Rightarrow2x^3=16y^3\\ \Rightarrow x^3=8y^3\\ \Rightarrow x=2y\)

Mà \(x^6\cdot y^6=64\Rightarrow\left(2y\right)^6\cdot y^6=64\Rightarrow64\cdot y^{12}=64\)

\(\Rightarrow y^{12}=1\Rightarrow\left[{}\begin{matrix}y=1\Rightarrow x=2\\y=-1\Rightarrow x=-2\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(2;1\right);\left(-2;-1\right)\)