Câu hỏi:

Ta có \({{(a+b+c)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca\)

Theo bài, \({{(a+b+c)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}\) Nên suy ra \(ab+bc+ca=0\).

Đặt \(x=ab;y=bc;z=ca\).

Khi đó \(x+y+z=0\). Suy ra \(x+y=-z;y+z=-x;z+x=-y\).

Xét \(\left( 1+\frac{a}{c} \right)\left( 1+\frac{b}{a} \right)\left( 1+\frac{c}{b} \right)\)

\(=\left( 1+\frac{ab}{bc} \right)\left( 1+\frac{bc}{ca} \right)\left( 1+\frac{ca}{ab} \right)\)

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

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

\(=\frac{-z}{y}\cdot \frac{-x}{z}\cdot \frac{-y}{x}=-1\)

Xét \(\frac{{{a}^{3}}{{b}^{3}}+{{b}^{3}}{{c}^{3}}+{{c}^{3}}{{a}^{3}}}{3{{a}^{2}}{{b}^{2}}{{c}^{2}}}=\frac{{{x}^{3}}+{{y}^{3}}+{{z}^{3}}}{3xyz}\)

\(=\frac{{{(x+y)}^{3}}-3xy(x+y)+{{z}^{3}}}{3xyz}\)

\(=\frac{{{(-z)}^{3}}-3xy(-z)+{{z}^{3}}}{3xyz}\)

\(=\frac{-{{z}^{3}}+3xyz+{{z}^{3}}}{3xyz}=\frac{3xyz}{3xyz}=1\) 

Từ đó, \(T=\frac{{{a}^{3}}{{b}^{3}}+{{b}^{3}}{{c}^{3}}+{{c}^{3}}{{a}^{3}}}{3{{a}^{2}}{{b}^{2}}{{c}^{2}}}+\left( 1+\frac{a}{c} \right)\left( 1+\frac{b}{a} \right)\left( 1+\frac{c}{b} \right)=1+(-1)=0\).

Vậy \(T=0\).

Có ba đa thức là: \(x^2+y^2 ; 2025 ; \frac{x}{2}+x y z\).

\(\begin{align}& (x+2)^3-(x-2)^3 \\& =x^3+6 x^2+12 x+8-\left(x^3-6 x^2+12 x-8\right) \\& =x^3+6 x^2+12 x+8-x^3+6 x^2-12 x+8 \\& =\left(x^3-x^3\right)+\left(6 x^2+6 x^2\right)+(12 x-12 x)+(8+8) \\& =12 x^2+16.\end{align}\)

- \(x(x+1) y\) chia hết cho \(x, x(x+1) y\) không chia hết cho \(y^2\)