Câu hỏi:

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28%.

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

Theo đề 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.

Ta có: \({{x}^{2}}+{{y}^{2}}+{{z}^{2}}=xy+yz+xz\)

\(2\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)=2(xy+yz+xz)\)

\(2{{x}^{2}}+2{{y}^{2}}+2{{z}^{2}}-2xy-2yz-2xz=0\)

\(\left( {{x}^{2}}-2xy+{{y}^{2}} \right)+\left( {{y}^{2}}-2yz+{{z}^{2}} \right)+\left( {{z}^{2}}-2xz+{{x}^{2}} \right)=0\)

\({{(x-y)}^{2}}+{{(y-z)}^{2}}+{{(z-x)}^{2}}=0\)

Vì \({{(x-y)}^{2}}\ge 0;{{(y-z)}^{2}}\ge 0;{{(z-x)}^{2}}\ge 0  \forall x,y,z\in \mathbb{R}\)

Nên \({{(x-y)}^{2}}+{{(y-z)}^{2}}+{{(z-x)}^{2}}\ge 0   \forall x,y,z\in \mathbb{R}\).

Để \({{(x-y)}^{2}}+{{(y-z)}^{2}}+{{(z-x)}^{2}}=0\) thì \(\left\{ \begin{array}{*{35}{l}} {{(x-y)}^{2}}=0 \\ {{(y-z)}^{2}}=0 \\ {{(z-x)}^{2}}=0 \\ \end{array} \right.\).

Tức là \(\left\{ \begin{array}{*{35}{l}} x=y \\ y=z \\ z=x \\ \end{array} \right.\) hay \(x=y=z\).

Từ đó ta có \(x+y+z=-3\). và \(x=y=z\).

Suy ra \(x=y=z=-1\).

Do đó \(B={{x}^{2022}}+{{y}^{2023}}+{{z}^{2024}}\) \(={{(-1)}^{2022}}+{{(-1)}^{2023}}+{{(-1)}^{2024}}\) \(=1+(-1)+1=-1\).

Vậy \(B={{x}^{2022}}+{{y}^{2023}}+{{z}^{2024}}=-1\).

\({{x}^{3}}+{{y}^{3}}=\left( x+y \right)\left( {{x}^{2}}-xy+{{y}^{2}} \right)=\left( x+y \right)\left( {{x}^{2}}+2xy+{{y}^{2}}-3xy \right)\)

\(=\left( x+y \right)\left[ \left( {{x}^{2}}+2xy+{{y}^{2}} \right)-3xy \right]=\left( x+y \right)\left[ {{(x+y)}^{2}}-3xy \right]\)

Với \(x+y=4\) và \(xy=3\) ta có \({{x}^{3}}+{{y}^{3}}=4\left( {{4}^{2}}-3.3 \right)=28\)