Giải:

Từ \(a^3+b^3+c^3=3abc\Leftrightarrow\)\(\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)

Ta xét các trường hợp:

Trường hợp \(1\): Nếu \(a+b+c=0\) thì:

\(\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\a+c=-b\end{matrix}\right.\)

Thay vào \(P\) ta có:

\(P=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)

\(=\left(\dfrac{a+b}{b}\right)\left(\dfrac{b+c}{c}\right)\left(\dfrac{a+c}{c}\right)\)

\(=\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}=\dfrac{\cdot\left(-c\right).\left(-a\right).\left(-b\right)}{b.c.a}=-1\)

Trường hợp \(2\): Nếu \(a=b=c\) thì:

\(P=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)

\(=\left(1+\dfrac{a}{a}\right)\left(1+\dfrac{a}{a}\right)\left(1+\dfrac{a}{a}\right)\)

\(=\left(1+1\right)\left(1+1\right)\left(1+1\right)\)

\(=2.2.2=8\)

Vậy \(P=-1\) hoặc \(P=8\)

Ta có:

\(S=1+3+3^2+3^3+...+3^{100}\)

\(=\left(1+3+3^2+3^3+3^4\right)+...+\left(3^{96}+3^{97}+3^{98}+3^{99}+3^{100}\right)\)

\(=1\left(1+3+3^2+3^3+3^4\right)+...+3^{96}\left(1+3+3^2+3^3+3^4\right)\)

\(=1.121+...+3^{96}.121\)

\(=121\left(1+...+3^{96}\right)⋮121\)

Vậy \(S\div121\) có chữ số tận cùng là \(0\)

Theo đề bài ta có:

\(3xy-5=x^2+2y\)

\(\Rightarrow3xy-2y=x^2+5\)

\(\Rightarrow y\left(3x-2\right)=x^2+5\left(1\right)\)

Do \(x,y\) là các số nguyên

Nên \(x^2+5⋮3x-2\Rightarrow9\left(x^2+5\right)⋮3x-2\)

\(\Rightarrow9x^2+45⋮3x-2\)

\(\Rightarrow9x^2-6x+6x-4+49⋮3x-2\)

\(\Rightarrow3x\left(3x-2\right)+2\left(3x-2\right)+49⋮3x-2\)

\(\Rightarrow49⋮3x-2\Rightarrow3x-2\inƯ\left(49\right)\)

\(\Rightarrow3x-2\in\left\{\pm1;\pm7;\pm49\right\}\Rightarrow x=\left\{1;3;17\right\}\)

Thay lần lượt vào \(\left(1\right)\Rightarrow y\in\left\{6;2;6\right\}\)

Vậy các cặp số \(\left(x,y\right)=\left(1;6\right);\left(3;2\right);\left(17;6\right)\)