Question

[Nguyễn Hoàng Anh Thư]

Cho \(a^3+b^3+c^3=3abc\) với a,b,c khác 0 

tính giá trị biểu thức 
P=\(\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)+\left(1+\frac{c}{a}\right)\)

Answer 1

[ST]

Sửa đề: tính P=(1+a/b)(1+b/c)(1+c/a)

\(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\left(1\right)\\a^2+b^2+c^2-ab-bc-ca=0\left(2\right)\end{cases}}\)

- Xét (1) ta có: \(a+b+c=0\Leftrightarrow\hept{\begin{cases}-a=b+c\\-b=c+a\\-c=a+b\end{cases}}\)

=> \(P=\frac{a+b}{b}\cdot\frac{b+c}{c}\cdot\frac{c+a}{a}=\frac{\left(-c\right).\left(-a\right).\left(-b\right)}{bca}=-\frac{abc}{abc}=-1\)

- Xét (2) ta có: \(a^2+b^2+c^2-ab-bc-ca=0\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

Mà \(\hept{\begin{cases}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{cases}\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0}\)

\(\Rightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Rightarrow a=b=c}\)

=>\(P=\frac{a+b}{b}\cdot\frac{b+c}{c}\cdot\frac{c+a}{a}=\frac{2a}{a}\cdot\frac{2a}{a}\cdot\frac{2a}{a}=2.2.2=8\)

Vậy P=-1 hoặc P=8

Answer 2

[tth_new]

Ta có; \(a^3+b^3+c^3=3abc\) hay \(a^3+b^3+c^3-3abc=0\)

Suy ra \(a+b+c=0\) hoặc a =  b = c. (bạn tự chứng minh)

* Nếu a + b + c = 0 thì:

\(P=\frac{a+b}{b}.\frac{b+c}{c}.\frac{c+a}{a}=\frac{-c}{b}.\frac{-a}{c}.\frac{-b}{a}=-1\)

*Nếu a = b  = c thì \(P=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)