Question

\(\left\{{}\begin{matrix}x-\frac{1}{x^3}=y-\frac{1}{y^3}\\\left(x-4y\right)\left(2x-y+\text{4}\right)=-36\end{matrix}\right.\)

Answer 1

ĐKXĐ: ...

\(\left(x-4y\right)\left(2x-y+4\right)=-36\)

\(\Leftrightarrow2x^2+4x-9xy+4y^2-16y=-36\)

\(\Leftrightarrow xy=\frac{2\left(x+1\right)^2+4\left(y-2\right)^2}{9}+2>0\)

\(x-y+\frac{x^3-y^3}{x^3y^3}=0\)

\(\Leftrightarrow x-y+\frac{\left(x-y\right)\left(x^2+xy+y^2\right)}{x^3y^3}=0\)

\(\Leftrightarrow\left(x-y\right)\left(1+\frac{x^2+xy+y^2}{x^3y^3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=y\\1+\frac{x^2+xy+y^2}{x^3y^3}=0\left(1\right)\end{matrix}\right.\)

Do \(xy>0\Rightarrow1+\frac{x^2+xy+y^2}{x^3y^3}>0\) \(\Rightarrow\left(1\right)\) vô nghiệm

Vậy \(x=y\)

\(\Rightarrow\left(x-4x\right)\left(2x-x+4\right)=-36\Leftrightarrow...\)