Giải thích các bước giải:

Ta có:

\(ab = cd \Leftrightarrow \dfrac{a}{c} = \dfrac{d}{b}\) 

Đặt \(\dfrac{a}{c} = \dfrac{d}{b} = k \Rightarrow \left\{ \begin{array}{l}
a = k.c\\
d = k.b
\end{array} \right.\)

Ta có:

\(\begin{array}{l}
{a^{2020}} + {b^{2020}} + {c^{2020}} + {d^{2020}}\\
 = {\left( {k.c} \right)^{2020}} + {b^{2020}} + {c^{2020}} + {\left( {k.b} \right)^{2020}}\\
 = {k^{2020}}.{c^{2020}} + {b^{2020}} + {c^{2020}} + {k^{2020}}.{b^{2020}}\\
 = {k^{2020}}.\left( {{c^{2020}} + {b^{2020}}} \right) + \left( {{b^{2020}} + {c^{2020}}} \right)\\
 = \left( {{b^{2020}} + {c^{2020}}} \right).\left( {{k^{2020}} + 1} \right)
\end{array}\)

Suy ra tổng \({a^{2020}} + {b^{2020}} + {c^{2020}} + {d^{2020}}\) đã cho là một hợp số.

Ta có:

ab=cd⇔ac=dbab=cd⇔ac=db 

Đặt ac=db=k⇒{a=k.cd=k.bac=db=k⇒{a=k.cd=k.b

Ta có:

a^2020+b^2020+c^2020+d^2020=(k.c)^2020+b^2020+c^2020+(k.b)^2020=k^2020.c^2020+b^2020+c^2020+k^2020.b^2020=k^2020.(c^2020+b^2020)+(b^2020+c^2020)=(b^2020+c^2020).(k^2020+1).a^2020+b^2020+c^2020+d^2020=(k.c)^2020+b^2020+c^2020+(k.b)^2020=k^2020.c^2020+b^2020+c^2020+k^2020.b^2020=k^2020.(c^2020+b^2020)+(b^2020+c^2020)=(b^2020+c^2020).(k^2020+1)

=) Tổng a^2020+b^2020+c^2020+d^2020 là hợp số.

Cho mình 5 sao và ctlhn nha thanks!!

@toyeucau_20