Question

tìm các nghiệm nguyên của phương trình

\(\frac{1}{x^{2^{ }}\left(x^2+y^2\right)}+\frac{1}{\left(x^2+y^2\right)\left(x^2+y^2+z^2\right)}+\frac{1}{x^2\left(x^2+y^2+z^2\right)}\)=1

Answer 1

ĐKXĐ: ...

Đặt \(\left(x^2;x^2+y^2;x^2+y^2+z^2\right)=\left(a;b;c\right)\Rightarrow1\le a\le b\le c\)

\(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\Leftrightarrow a+b+c=abc\)

\(\Rightarrow abc\le3c\Rightarrow ab\le3\Rightarrow ab=\left\{1;2;3\right\}\)

- TH1: \(ab=1\Rightarrow a=b=1\Rightarrow2+c=c\) (vô nghiệm)

- TH2: \(ab=2\Rightarrow\left\{{}\begin{matrix}a=1\\b=2\end{matrix}\right.\) \(\Rightarrow c+3=2c\Rightarrow c=3\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=1\\x^2+y^2=2\\x^2+y^2+z^2=3\end{matrix}\right.\) \(\Rightarrow x^2=y^2=z^2=1\Rightarrow...\)

- TH3: \(ab=3\Rightarrow\left\{{}\begin{matrix}a=1\\b=3\end{matrix}\right.\) \(\Rightarrow c+4=3c\Rightarrow c=2< b\) (loại)

Vậy...