Question

[Ngọc Nhi]

tìm x, y

\(\hept{\begin{cases}\frac{y^2-x^2}{3}=\frac{y^2+x^2}{5}\\x^{10}y^{10}=1024\end{cases}}\)

Answer 1

[ST]

Ta có:\(\frac{y^2-x^2}{3}=\frac{y^2+x^2}{5}\Rightarrow5\left(y^2-x^2\right)=3\left(y^2+x^2\right)\Rightarrow5y^2-5x^2=3y^2+3x^2\Rightarrow2y^2=8x^2\Rightarrow y^2=4x^2\)

\(\Rightarrow\frac{y^2}{4}=\frac{x^2}{1}\Rightarrow\frac{y}{2}=\frac{x}{1}\)

Đặt \(\frac{x}{1}=\frac{y}{2}=k\Rightarrow x=k,y=2k\)

Lại có: \(x^{10}y^{10}=k^{10}.\left(2k\right)^{10}=k^{10}.1024k^{10}=1024k^{20}=1024\)

\(\Rightarrow k^{20}=1\Rightarrow k=\pm1\)

Với k = 1 => x = 1, y = 2

Với k = -1 => x = -1, y = -2

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