Gọi \(M=\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\)      ta có :

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

\(=1+\frac{c}{a-b}.\frac{b^2-bc+ac-a^2}{ab}\)

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

\(=1+\frac{2c^2}{ab}=1+\frac{2c^3}{abc}\)

Tương tự  \(M.\frac{a}{b-c}=1+\frac{2a^3}{abc}\)

và  \(M.\frac{b}{c-a}=1+\frac{2b^3}{abc}\)

Vậy \(A=3+\frac{2\left(a^3+b^3+c^3\right)}{abc}=9\)

( Vì \(a^3+b^3+c^3=3abc\).  Lại do  . ( Phân tích là ra hết ).\(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

=> ....

bài này trong sách nâng cao phát triển tập 1 

Cách 1 . \(A=\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right)\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)\)

Đặt \(\frac{a-b}{c}=x\); \(\frac{b-c}{a}=y\) ; \(\frac{c-a}{b}=z\)

Ta có : \(\frac{x+y}{z}=\frac{\frac{a-b}{c}+\frac{b-c}{a}}{\frac{c-a}{b}}=\frac{ab\left(a-b\right)+cb\left(b-c\right)}{ac\left(c-a\right)}=\frac{b\left(b-a-c\right)}{ac}=\frac{2b^2}{ac}=\frac{2b^3}{abc}\)

tương tự : \(\frac{y+z}{x}=\frac{2c^3}{abc}\); \(\frac{x+z}{y}=\frac{2a^3}{abc}\)

\(\Rightarrow A=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=1+\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+1+\frac{y}{z}+\frac{z}{x}+\frac{z}{y}+1\)

\(=3+\frac{x+z}{y}+\frac{x+y}{z}+\frac{y+z}{x}=3+\frac{2}{abc}\left(a^3+b^3+c^3\right)\)

Áp dụng bài toán phụ : Nếu a + b + c = 0 thì \(a^3+b^3+c^3=3abc\) (có thể chứng minh bằng cách rút a = - b - c  rồi thay vào tổng ba lập phương) được : 

\(A=3+\frac{2}{abc}.3abc=3+6=9\)

Đặt \(\frac{a-b}{c}=x=>\frac{c}{a-b}=\frac{1}{x}\)

\(\frac{b-c}{a}=y=>\frac{a}{b-c}=y\)

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

=>\(A=\left(x+y+z\right).\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

=>\(A=x.\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+y.\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+z.\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

=>\(A=1+\frac{x}{y}+\frac{x}{z}+1+\frac{y}{x}+\frac{y}{z}+1+\frac{z}{x}+\frac{z}{y}\)

=>\(A=3+\left(\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+\frac{y}{z}+\frac{z}{x}+\frac{z}{y}\right)\)

=>\(A=3+\frac{x+z}{y}+\frac{x+y}{z}+\frac{y+z}{x}\)

Lại có: \(\frac{x+z}{y}=\frac{\frac{a-b}{c}+\frac{c-a}{b}}{\frac{b-c}{a}}=\frac{\frac{ab-b^2}{bc}+\frac{c^2-ac}{bc}}{\frac{b-c}{a}}=\frac{\frac{ab-b^2+c^2-ac}{bc}}{\frac{b-c}{a}}\)

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

\(=\frac{\left(a-b-c\right).\left(b-c\right).a}{\left(b-c\right).bc}=\frac{\left(a-b-c\right).a}{bc}=\frac{\left(a+a-a-b-c\right).a}{bc}\)

\(=\frac{\left[2a-\left(a+b+c\right)\right].a}{bc}\)

Vì a+b+c=0

=>\(\frac{x+z}{y}=\frac{\left(2a-0\right).a}{bc}=\frac{2a^2}{bc}=\frac{2a^3}{abc}\)

Chứng minh tương tự, ta có:

\(\frac{x+y}{z}=\frac{2b^3}{abc}\)

\(\frac{y+z}{x}=\frac{2c^3}{abc}\)

=>\(A=3+\frac{x+z}{y}+\frac{x+y}{z}+\frac{y+z}{x}=3+\frac{3a^3}{abc}+\frac{3b^3}{abc}+\frac{3c^3}{abc}\)

=>\(A=3+\frac{2a^3+2b^3+2c^3}{abc}\)

=>\(A=3+\frac{2.\left(a^3+b^3+c^3\right)}{abc}\)

Vì a+b+c=0

=>a=-(b+c)

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

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

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

=>\(a^3=-b^3-3bc.\left(b+c\right)-c^3\)

=>\(a^3+b^3+c^3=-3bc.\left(b+c\right)\)

Vì a+b+c=0=>b+c=-a

=>\(a^3+b^3+c^3=-3bc.\left(-a\right)\)

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

Thay vào A, ta có:

\(A=3+\frac{2.\left(a^3+b^3+c^3\right)}{abc}=3+\frac{2.3abc}{abc}=3+\frac{6.abc}{abc}=3+6=9\)

=>A=9

Vậy A=9

Cách 2. Đặt \(P=\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\) ; \(Q=\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\)

\(\Rightarrow P=\frac{ab\left(a-b\right)+bc\left(b-c\right)+ac\left(c-a\right)}{abc}\)

Xét riêng : \(ab\left(a-b\right)+bc\left(b-c\right)+ac\left(c-a\right)=ab\left[-\left(b-c\right)-\left(c-a\right)\right]+bc\left(b-c\right)+ac\left(c-a\right)\)

\(=\left[-ab\left(b-c\right)+bc\left(b-c\right)\right]+\left[-ab\left(c-a\right)+ac\left(c-a\right)\right]\)

\(=b.\left(c-a\right).\left(b-c\right)+a\left(c-a\right)\left(c-b\right)=\left(c-a\right)\left(b-c\right)\left(b-a\right)\)

Vậy : \(P=\frac{\left(c-a\right)\left(b-c\right)\left(b-a\right)}{abc}\)

Tiếp theo, rút gọn Q như sau : 

Đặt \(x=b-c\); \(y=c-a\); \(z=a-b\)

Ta có : \(x-y=a+b-2c=-c-2c=-3c\)

\(y-z=b+c-2a=-a-2a=-3a\)

\(z-x=c+a-2b=-b-2b=-3b\)

\(\Rightarrow3Q=\frac{-\left(y-z\right)}{x}+\frac{-\left(z-x\right)}{y}+\frac{-\left(x-y\right)}{z}\)\(\Rightarrow-3Q=\frac{y-z}{x}+\frac{z-x}{y}+\frac{x-y}{z}\)

Rút gọn tương tự như P, ta được : \(-3Q=\frac{\left(x-y\right)\left(y-z\right)\left(x-z\right)}{xyz}=\frac{\left(-3c\right).\left(-3a\right).\left(3b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

\(\Rightarrow Q=-\frac{9abc}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

Vậy : \(A=PQ=\frac{\left(c-a\right)\left(c-b\right)\left(a-b\right)}{abc}.\frac{-9abc}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)\(\Rightarrow A=9\)

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 ^^

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

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

\(=1+\frac{c}{a-b}.\frac{b^2-bc+ca-a^2}{ab}\)

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

Tương tự : \(M.\frac{a}{b-c}=1+\frac{2a^2}{bc};M.\frac{b}{c-a}=1+\frac{2b^2}{ca}\)

Do vậy \(A=3+2.\frac{a^3+b^3+c^3}{abc}=9\left(do.a+b+c=0.thi.a^3+b^3+c^3=3abc\right)\)

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

\(\Rightarrow\hept{\begin{cases}b+c-a=ck\\a+b+c=bk\\b-c+a=ak\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}2b=k\left(a+c\right)\left(1\right)\\2c=k\left(b-a\right)\left(2\right)\\2b+2c=b\left(b+c\right)\Rightarrow k=2\end{cases}}\)

Thay k=2 vào (1) và (2) : 

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

Vậy \(\frac{\left(b-a\right)\left(c+b\right)\left(a+c\right)}{abc}=\frac{\left(b-a\right)\left(c+b\right)\left(a+c\right)}{\left(b-c\right)\left(a+c\right)\left(b-a\right)}=\frac{b+c}{b-c}\)

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

Ta có:\(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)

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

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

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

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

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

<=> \(A=0\)

=> ....

đề bài thiếu nhé  , a,b,c khác nhau nhé :)

có :\(a=b-c\)

vì a,b,c khác nhau

\(\Rightarrow b-c\ne0\)

có:

\(a+b+c=0\Leftrightarrow c=a-b.\)

\(a=b-c\)

\(b=c-a\)

thày vào M ta được

\(\left(\frac{c}{c}+\frac{a}{a}+\frac{b}{b}\right)\left(\frac{c}{c}+\frac{a}{a}+\frac{b}{b}\right)=9\)

pain sai r nhé

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

Đặt \(\left(\frac{a-b}{c};\frac{b-c}{a};\frac{c-a}{b}\right)\rightarrow\left(x;y;z\right)\)

Khi đó \(M=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

\(=3+\frac{x+y}{z}+\frac{y+z}{x}+\frac{x+z}{y}\)

Ta có:\(\frac{x+y}{z}=\left(\frac{a-b}{c}+\frac{b-c}{a}\right)\cdot\frac{b}{c-a}\)

\(=\frac{a^2-ab+bc-c^2}{ca}\cdot\frac{b}{c-a}\)

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

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

\(=\frac{-\left(a-b+c\right)\cdot b}{ca}=\frac{-\left(a+b+c-2b\right)\cdot b}{ca}=\frac{2b^3}{abc}\)

Tương tự ta có:

\(\frac{y+z}{x}=\frac{2c^3}{abc};\frac{x+z}{y}=\frac{2a^3}{abc}\)

Khi đó:\(M=\frac{2a^3+2b^3+2c^3}{abc}+3\)

Mặt khác:\(a+b+c=0\Rightarrow a+b=-c\Rightarrow a^3+3ab\left(a+b\right)+b^3=-c^3\)

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

Khi đó \(M=\frac{6abc}{abc}+3=9\)

Áp dụng BĐT Bunhiacopxki:

\(\sqrt{\left(a+b\right)\left(a+c\right)}\ge\sqrt{ac}+\sqrt{ab}\)

\(\Rightarrow\)\(\frac{a}{a+\sqrt{\left(a+b\right)\left(a+c\right)}}\)\(\le\frac{a}{a+\sqrt{ab}+\sqrt{ac}}\)=\(\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)(1)

Tương tự ta có: \(\frac{b}{b+\sqrt{\left(b+c\right)\left(b+a\right)}}\le\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)(2)

\(\frac{c}{c+\sqrt{\left(c+a\right)\left(c+b\right)}}\le\frac{\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)(3)

Cộng theo vế của (1);(2)&(3) ta đc:

A\(\le1\)

Dấu''='' xảy ra\(\Leftrightarrow\)a=b=c