Do \(a^3+b^3+c^3=3abc\)
\(\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
Mà \(a+b+c\ne0\)
\(\Rightarrow a^2+b^2+c^2-ab-bc-ac=0\)
\(\Rightarrow a^2+b^2+c^2=ab+bc+ac\)
Khi đó:
\(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ac\right)\)\(=\left(a^2+b^2+c^2\right)+2\left(a^2+b^2+c^2\right)=3\left(a^2+b^2+c^2\right)\)
Vậy: \(N=\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\frac{a^2+b^2+c^2}{3\left(a^2+b^2+c^2\right)}=\frac{1}{3}\)

1) Thay xyz = 1  , ta có : 

 \(\frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+xz}=\frac{z}{z+xz+xyz}+\frac{xz}{xz+xyz+xyz^2}+\frac{1}{1+z+xz}\)

\(=\frac{z}{z+xz+1}+\frac{xz}{xz+1+z}+\frac{1}{z+xz+1}=\frac{z+xz+1}{z+xz+1}=1\)

2) Phân tích A thành nhân tử được \(A=\left(a-b\right)\left(b-c\right)\left(a-c\right)\left(a+b+c\right)\)

Vì a + b + c = 0 nên A = 0

3) Phân tích  A thành  \(\frac{\left(b-a\right)\left(a-c\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=1\)

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 ^^

\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b+b+c+c+a}{c+a+b}=2\)(T/C...)

Xét a+b+c=0

\(\Rightarrow a+b=-c,c+b=-a,a+c=-b\)

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

Xét a+b+c\(\ne0\)

\(\Rightarrow a+b=2c,b+c=2a,c+a=2b\)

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

Giải:
+) Xét a + b + c = 0

\(\Rightarrow-a=b+c\)

\(\Rightarrow-b=a+c\)

\(\Rightarrow-c=a+b\)

Ta có:

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

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

+) Xét \(a+b+c\ne0\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b+b+c+c+a}{a+b+c}=\frac{2a+2b+2c}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)

Ta có:

\(M=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}.\frac{b+c}{c}.\frac{a+c}{a}=\frac{a+b}{c}.\frac{b+c}{a}.\frac{c+a}{b}=2.2.2=8\)

Vậy M = -1 hoặc M = 8

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

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

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

không hỉu

chỉnh lại rồi nhé

18. Ta có : \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\Rightarrow\frac{ayz+bxz+cxy}{xyz}=0\Rightarrow ayz+bxz+cxy=0\)

\(\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac}\right)=1\)

\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2xyz\left(\frac{1}{abz}+\frac{1}{xbc}+\frac{1}{acy}\right)=1\)

\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2xyz\left(\frac{ayz+bxz+cxy}{abcxyz}\right)=1\)

\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\)

19. Nhân cả hai vế của đẳng thức giả thiết với \(\frac{1}{b-c}+\frac{1}{c-a}+\frac{1}{a-b}\)được 

\(\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\left(\frac{1}{b-c}+\frac{1}{c-a}+\frac{1}{a-b}\right)=0\)

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

Ta có ;

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

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

17. Xét vế trái ; 

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

\(=\frac{a}{-a\left(b^2+b+1\right)}+\frac{b}{-b\left(a^2+a+1\right)}=\frac{-1}{b^2+b+1}+\frac{-1}{a^2+a+1}\)

\(=\frac{-\left(a^2+a+1+b^2+b+1\right)}{\left(a^2+a+1\right)\left(b^2+b+1\right)}=\frac{-\left[\left(a+b\right)^2-2ab+3\right]}{a^2b^2+ab\left(a+b\right)+a^2+b^2+ab+2}\)\(=\frac{2\left(ab-2\right)}{a^2b^2+\left(a^2+2ab+b^2\right)+2}=\frac{2\left(ab-2\right)}{a^2b^2+3}\)