1)1-1-2

2)2-1-3

3)2-6-2-3

4)1-6-2-3

5)1-3-1-3

6)1-1-1-1

7)2-2-3

8)4-5-2

9)1-6-2-3

1)1+1->2

2)2->1+3

3)2+6->2+3

158cm³

\(A=\frac{1}{x^2-x}+\frac{1}{x^2+x+1}+\frac{2x}{1-x^3}\)

\(A=\frac{1}{x.\left(x-1\right)}+\frac{1}{x^2+x+1}+\frac{2x}{\left(1-x\right)\left(x^2+x+1\right)}\)

\(A=\frac{x^2+x+1}{x.\left(x-1\right)\left(x^2+x+1\right)}+\frac{x\left(x-1\right)}{x.\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x^2}{x.\left(x-1\right)\left(x^2+x+1\right)}\)

\(A=\frac{x^2+x+1}{x.\left(x-1\right)\left(x^2+x+1\right)}+\frac{x^2-x}{x.\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x^2}{x.\left(x-1\right)\left(x^2+x+1\right)}\)

\(A=\frac{x^2+x+1+x^2-x-2x^2}{x.\left(x-1\right)\left(x^2+x+1\right)}\)

\(A=\frac{1}{x.\left(x-1\right)\left(x^2+x+1\right)}\)

\(A=\frac{1}{x.\left(x^3-1\right)}\)

Với x=10

\(\Rightarrow A=\frac{1}{10.\left(10^3-1\right)}\)

\(A=\frac{1}{10.999}\)

\(A=\frac{1}{9990}\)

Vậy \(A=\frac{1}{9990}\)tại x=10

\(\frac{ab+bc+ca}{abc}=0\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)

\(A=\frac{bc}{a^2}+\frac{ca}{b^2}+\frac{ab}{c^2}=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=abc.\frac{3}{abc}=3\)

Pham Van Hung mình ko hiểu tại sao \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)