Rozwiąż w liczbach rzeczywistych:

[karol2859]

Rozwiąż w liczbach rzeczywistych:
\(\displaystyle{ 1. \begin{cases} x ^{5}-y ^{5} =932 \\ x-y=2 \end{cases}}\)
Rozwiąż w liczbach naturalnych:
\(\displaystyle{ 2. \begin{cases} x ^{2}+y-z=100 \\ x+y ^{2} -z=124 \end{cases}}\)
Proszę o pomoc:)

[cienkibolek]

\(\displaystyle{ \begin{cases}
x^{5}-y^{5}=932\\
x-y=2
\end{cases}
\\
(x-y)^{5}=x^{5}-5x^{4}y+10x^{3}y^{2}-10x^{2}y^{3}+5xy^{4}-y^{5}\\
32=x^{5}-y^{5}-5x^{4}y+10x^{3}y^{2}-10x^{2}y^{3}+5xy^{4}\\
32=932-5x^{4}y+10x^{3}y^{2}-10x^{2}y^{3}+5xy^{4}\\
5x^{4}y-10x^{3}y^{2}+10x^{2}y^{3}-5xy^{4}=900\\
x^{4}y-2x^{3}y^{2}+2x^{2}y^{3}-xy^{4}=180\\
x^{4}y-xy^{4}-2x^{3}y^{2}+2x^{2}y^{3}=180\\
x^{4}y-xy^{4}-2x^{2}y^{2}(x-y)=180\\
x^{4}y-xy^{4}-4x^{2}y^{2}=180}\)

\(\displaystyle{ (x-y)(x^{3}y+xy^{3})=x^{4}y+x^{2}y^{3}-x^{3}y^{2}-xy^{4}=\\
=x^{4}y-xy^{4}+x^{2}y^{3}-x^{3}y^{2}=\\
=x^{4}y-xy^{4}-x^{2}y^{2}(x-y)=\\
=x^{4}y-xy^{4}-2x^{2}y^{2}\\
\\
x^{4}y-xy^{4}-2x^{2}y^{2}=2(x^{3}y+xy^{3})\\
x^{4}y-xy^{4}=2(x^{3}y+xy^{3})+2x^{2}y^{2}\\
\\
2(x^{3}y+xy^{3})+2x^{2}y^{2}-4x^{2}y^{2}=180\\
2(x^{3}y+xy^{3})-2x^{2}y^{2}=180\\
x^{3}y+xy^{3}-x^{2}y^{2}=90\\
\\
(x-y)(x^{2}y-xy^{2})=x^{3}y-x^{2}y^{2}-x^{2}y^{2}+xy^{3}\\
2(x^{2}y-xy^{2})=x^{3}y+xy^{3}-x^{2}y^{2}-x^{2}y^{2}\\
\\
2(x^{2}y-xy^{2})=90-x^{2}y^{2}\\
\\
2xy(x-y)=2x^{2}y-2xy^{2}\\
\\
4xy+x^{2}y^{2}=90\\
(xy)^{2}+4xy-90=0}\)

\(\displaystyle{ xy=-2 \pm \sqrt{94} \\
\\
\begin{cases}
xy=-2 \pm \sqrt{94} \\
x-y=2
\end{cases}
\\
y=x-2\\
x(x-2)=-2 \pm \sqrt{94}\\
x^{2}-2x+2- \sqrt{94}=0 \vee x^{2}-2x+2+\sqrt{94} \neq 0 (\Delta<0)\\
\\
x^{2}-2x+2- \sqrt{94}=0\\
\Delta=4( \sqrt{94}-1 )\\
\\
x_{1}=1+ \sqrt{ \sqrt{94} -1} \vee x_{2}=1- \sqrt{ \sqrt{94} -1}\\
y_{1}= \sqrt{ \sqrt{94} -1}-1 \vee y_{2}=- \sqrt{ \sqrt{94} -1}-1}\)

co do drugiego to metodą prób i błędów wyszło mi:
\(\displaystyle{ x=12\\
y=13\\
z=57\\}\)
prawdopodobnie jedyne rozwiązanie