Question

[Lê Phương Thảo]

\(\frac{x^3+y^3}{6}=\frac{x^3-2y^3}{4}\) va \(x^6.y^6=64\)

cho mk may like di may ban

Answer 1

[kaitovskudo]

Bạn nên tính x⁶.y⁶=64     =>(xy)⁶=2⁶

=>xy=2 sẽ dễ tính hơn

Answer 2

[Lê Phương Thảo]

\(\frac{x^3+y^3}{6}=\frac{x^3-2y^3}{4}=\frac{2x^3+2y^3}{12}=\frac{2x^3+2y^3+x^3-2y^3}{12+4}=\frac{3x^3}{16}\)

\(\frac{x^3+y^3}{6}=\frac{x^3-2y^3}{4}=\frac{\left(x^3+y^3\right)-\left(x^3-2y^3\right)}{6-4}=\frac{3y^3}{2}\)

Do do : \(\frac{3x^3}{16}=\frac{3y^3}{2}\Rightarrow\frac{x^3}{8}=\frac{y^3\Rightarrow x^6}{64}=y^6\)

Nhan moi tu so moi y⁶ ta duoc : \(\frac{x^6.y^6}{64}=y^{12}=\frac{64}{64}=1\Rightarrow y=+-1\&x=+-2\)

Vay ta co : (2;1) va (-2;-1)

tick mk nhe moi ng