\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=1\Rightarrow\frac{1}{a+b+c}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\Rightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b+c}-\frac{1}{c}\)

\(\Leftrightarrow\frac{a+b}{ab}=\frac{-a-b}{c\left(a+b+c\right)}\Rightarrow c\left(a+b\right)\left(a+b+c\right)=ab\left(-a-b\right)\)

\(\Rightarrow\left(a+b\right)\left(ca+cb+c^2\right)+ab\left(a+b\right)=0\Rightarrow\left(a+b\right)\left(ca+cb+c^2+ab\right)=0\)

\(\Rightarrow\left(a+b\right)\left(c+a\right)\left(b+c\right)=0\)

=> Trong 3 số a,b,c có 2 số đối nhau.Giả sử a = -b thì a⁹ + b⁹ = 0.

Tương tự giả sử b = -c hay a = -c thì b⁹⁹ + c⁹⁹ = 0 hay c⁹⁹⁹ + a⁹⁹⁹ = 0

Vậy biểu thức cần tính bằng 0.

bằng 0 quá dễ Hi Hi !!!

biểu thức trên = 0 đó

Ta có: \(a^3+b^3+c^3=3abc\)

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

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

\(\Leftrightarrow\left[\left(a+b\right)^3+c^3\right]-\left[3ab\left(a+b\right)+3abc\right]=0\)

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

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

Nếu \(a^2+b^2+c^2-ab-bc-ca=0\)

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

\(\Rightarrow a=b=c\)

Khi đó \(A=2^3=8\)

Nếu \(a+b+c=0\Rightarrow a+b=-c;b+c=-a;c+a=-b\)

Thay vào ta được:

\(A=\frac{a+b}{b}\cdot\frac{b+c}{c}\cdot\frac{c+a}{a}=\frac{-abc}{abc}=-1\)

Vậy A = 8 hoặc A = -1

Đề có chỗ bị lỗi kìa. Mai Mai

Ta có:

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

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

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

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

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

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

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

Vì a³+b³+c³=3abc và a+b+c khác 0

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

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

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

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

Tổng 3 số không âm = 0 <=> chúng đều = 0

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

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

\(\)

Ta có ; \(a^3+b^3+c^3=3abc\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)

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

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

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

Vì \(a+b+c\ne0\) nên ta có \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\Leftrightarrow a=b=c\)

a) Thay a = b = c vào biểu thức được : \(\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{3a^2}{9a^2}=\frac{1}{3}\)

b) Thay a = b = c vào P : \(P=\frac{2}{a}.\frac{2}{b}\frac{2}{c}=\frac{8}{abc}\)

b,

Ta có:  \(P=\left(\frac{1}{a}+\frac{1}{b}\right)\left(\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{c}+\frac{1}{a}\right)=\frac{a+b}{ab}.\frac{b+c}{bc}.\frac{a+c}{ca}\)

Vì a=b=c (chứng minh tương tự câu a)

=>a+b=2b

 b+c=2c

 a+c=2a

Do đó \(P=\frac{2b}{ab}.\frac{2c}{bc}.\frac{2a}{ca}=\frac{\left(2.2.2\right).abc}{\left(abc\right)^2}=\frac{8}{abc}\left(đpcm\right)\)

a) Ta có : \(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}\Leftrightarrow\frac{a+b}{c}+1=\frac{b+c}{a}+1=\frac{c+a}{b}+1\)

\(\Rightarrow\frac{a+b+c}{a}=\frac{a+b+c}{b}=\frac{a+b+c}{c}\)

\(\Rightarrow P=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)

b) Đề bài sai ^^

Ta có: a - b - c = 0

=> \(\hept{\begin{cases}a-c=b\\a-b=c\\-b-c=-a\end{cases}}\Rightarrow\hept{\begin{cases}a-c=b\\-\left(a-b\right)=-c\\-\left(b+c\right)=-a\end{cases}}\Rightarrow\hept{\begin{cases}a-c=b\\-a+b=-c\\b+c=a\end{cases}}\)

Lại có: \(P=\left(1-\frac{c}{a}\right)\left(1-\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\)

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

Ta có  \(\left(\frac{1}{a}+\frac{1}{b}\right)+\frac{1}{c}=\frac{1}{a+b+c}\) 

\(\Leftrightarrow\) \(\frac{a+b}{ab}=\frac{1}{a+b+c}-\frac{1}{c}=\frac{c-\left(a+b+c\right)}{c\left(a+b+c\right)}=\frac{-a-b}{c\left(a+b+c\right)}\)

\(\Leftrightarrow\) \(c\left(a+b\right)\left(a+b+c\right)+ab\left(a+b\right)=0\)

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

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

\(\Leftrightarrow\) \(\hept{\begin{cases}a+b=0\\b+c=0\\c+a\end{cases}}=0\Leftrightarrow\hept{\begin{cases}a=-b\\b=-c\\c=-a\end{cases}\Leftrightarrow\hept{\begin{cases}a^3=-b^3\\b^3=-c^3\\c^3=-a^3\end{cases}}\Leftrightarrow\hept{\begin{cases}a^3+b^3=0\\b^3+c^3=0\\c^3+a^3=0\end{cases}}}\) 

(ko có kí hiệu hoặc cho 3 cái nên mk dùng kí hiệu và nhé)

Do đó \(A=\left(a^3+b^3\right)\left(b^3+c^3\right)\left(c^3+a^3\right)=0\)

em mới học lớp 5 nên ko giúp đc gì, mong chị tha lỗi. chúc chị học giỏi nha

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

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

Vì \(a+b+c\ne0\Rightarrow a^2+b^2+c^2-ab-bc-ca=0\)

\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

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

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

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

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

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

\(\Rightarrow A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}\cdot\frac{b+c}{c}\cdot\frac{c+a}{a}=\frac{2a.2a.2a}{a.a.a}=\frac{8a^3}{a^3}=8\)