Đặt \(A=\left(\frac{a}{b}+1\right)\left(\frac{b}{c}+1\right)\left(\frac{c}{d}+1\right)\left(\frac{d}{a}+1\right)\)

\(\frac{-a+b+c+d}{a}=\frac{a-b+c+d}{b}=\frac{a+b-c+d}{c}=\frac{a+b+c-d}{d}=\frac{2\left(a+b+c+d\right)}{a+b+c+d}=2\)( tc dãy tỉ số bằng nhau )

\(\Rightarrow\hept{\begin{cases}-a+b+c+d=2a\\a-b+c+d=2b\\a+b-c+d=2c\end{cases}}\)và \(a+b+c-d=2d\)

\(\Rightarrow\hept{\begin{cases}a+b+c+d=4a\\a+b+c+d=4b\\a+b+c+d=4c\end{cases}}\)và \(a+b+c+d=4d\)

\(\Rightarrow4a=4b=4c=4d\)

\(\Rightarrow a=b=c=d\)thay vào bt A ta được:

\(A=2.2.2.2=16\)

Áp dụng tính chất của dãy tỉ số bằng nhau,ta có:

\(\frac{2a+b+c}{a}=\frac{2b+c+a}{b}=\frac{2c+a+b}{c}=\frac{2a+b+c+2b+c+a+2c+a+b}{a+b+c}=\frac{4\left(a+b+c\right)}{a+b+c}=4\)

\(\Rightarrow\frac{2a+b+c}{a}=4\Rightarrow2a+b+c=4a\Rightarrow b+c=4a-2a=2a\)

          \(\frac{2b+c+a}{b}=4\Rightarrow2b+c+a=4b\Rightarrow c+a=4b-2b=2b\)

          \(\frac{2c+a+b}{c}=4\Rightarrow2c+a+b=4c\Rightarrow a+b=4c-2c=2c\)   

Suy ra \(P=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\frac{2c.2a.2b}{abc}=\frac{8abc}{abc}=8\)

Vậy P=8

Cho hỏi tớ sai chỗ nào ạ :>?Góp ý giúp nha?

Ta có: a³+b³+c³=3abc <=> a³+b³+c³-3abc=0

<=>\(a^3+3a^2b+3ab^2+b^3+c^3-3ab\left(a+b\right)-3abc=0\)

<=>\(\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)

<=>\(\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)

<=>\(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)

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

Mà a+b+c khác 0

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

<=>\(2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

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

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

<=>\(\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Leftrightarrow}}a=b=c}\)

=>\(N=\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{3a^2}{9a^2}=\frac{1}{3}\)

- Ta có : \(a^3+b^3+c^3=3abc\)

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

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

Mà \(a+b+c\ne0\)

=> \(a^2+b^2+c^2-ab-bc-ac=0\)

=> \(\frac{\left(a^2-2ab+b^2\right)+\left(b^2-2ac+c^2\right)+\left(c^2-2ac+a^2\right)}{2}=0\)

=> \(\frac{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}{2}=0\)

=> \(a-b=b-c=c-a=0\)

=> \(a=b=c\)

- Thay a = b = c vào biểu thức N ta được :

\(N=\frac{a^2+a^2+a^2}{\left(a+a+a\right)^2}=\frac{3a^2}{9a^2}=\frac{1}{3}\)

Vậy giá trị của N = \(\frac{1}{3}\) khi \(a^3+b^3+c^3=3abc\) và \(a+b+c\ne0\)

pn oi nhieu the nay ai ma giai cho het dc

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