Ta có : \(\frac{b-c}{\left(a-b\right)\left(a-c\right)}=\frac{\left(a-c\right)-\left(a-b\right)}{\left(a-b\right)\left(a-c\right)}=\frac{a-c}{\left(a-b\right)\left(a-c\right)}-\frac{a-b}{\left(a-b\right)\left(a-c\right)}\)

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

Tương tự ta cũng chứng minh được :

\(\hept{\begin{cases}\frac{c-a}{\left(b-c\right)\left(b-a\right)}=\frac{1}{b-c}+\frac{1}{a-b}\left(2\right)\\\frac{a-b}{\left(c-a\right)\left(c-b\right)}=\frac{1}{c-a}+\frac{1}{b-c}\left(3\right)\end{cases}}\)

Từ (1), (2), (3), suy ra : \(\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-c\right)\left(b-a\right)}+\frac{a-b}{\left(c-a\right)\left(c-b\right)}\)

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

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

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

Tuong tu => DPCM

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

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

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

Cộng theo vế 3 đẳng thức trên ta có đpcm.

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

Khi đó :
\(Q=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=3+\frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y}\)

Ta có :

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

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

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

Hoàn toàn tương tự 

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

Do đó :

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

\(=3+\frac{2\left[\left(-c\right)^3-3ab\left(-c\right)^3+c^3\right]}{abc}=3+\frac{2.3abc}{abc}=3+6=9\)

Ta có đpcm

*Đặt P = (a-b)/c + (b-c)/a + (c-a)/b, ta có:
P = (a-b)/c + (b-c)/a + (c-a)/b
=> abc.P = ab(a-b) + bc(b-c) + ca(c-a)
= ab(a-b) + bc(b-a + a-c) + ca(c-a) 
= ab(a-b) - bc(a-b) - bc(c-a) + ca(c-a) 
= b(a-b)(a-c) + c(c-a)(a-b) 
= (a-b)(a-c)(b-c) 
=> P = (a-b)(a-c)(b-c)/abc 
*Đặt Q = c/(a-b) + a/(b-c) + b/(c-a), ta có:
Vì a+b+c = 0 => a+b = -c ; b+c = -a ; c+a = -b
Q = c/(a-b) + a/(b-c) + b/(c-a) 
=> (a-b)(b-c)(c-a).Q = c(b-c)(c-a) + a(a-b)(c-a) + b(a-b)(b-c) 
= c(b-c)(c-a) + (-b-c)(a-b)(c-a) + b(a-b)(b-c) 
= c(b-c)(c-a) – c(a-b)(c-a) – b(a-b)(c-a) + b(a-b)(b-c) 
= c(c-a)(2b-a-c) + b(a-b)(a+b-2c) 
= 3bc(c-a) – 3bc(a-b) 
= 3bc(b+c-2a) 
= 3bc(-a-2a) 
= -9abc 
=> Q = -9abc/(a-b)(b-c)(c-a) = 9abc /(a-b)(b-c)(a-c) 
Vậy P.Q = 9 (đpcm)

Đặt \(\frac{a-b}{c}=x;\frac{b-c}{a}=y;\frac{c-a}{b}=z\Rightarrow\frac{c}{a-b}=\frac{1}{x};\frac{a}{b-c}=\frac{1}{y};\frac{b}{c-a}=\frac{1}{z}\)

Vì a+b+c=0 => a=-b-c ; b=-c-a ; c=-a-b 

                         a³+b³+c³=3abc

Ta có: \(\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right)\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=3+\frac{x+z}{y}+\frac{x+y}{z}+\frac{y+z}{x}\)

Lại có: \(\frac{x+z}{y}=\left(x+z\right)\cdot\frac{1}{y}=\left(\frac{a-b}{c}+\frac{c-a}{b}\right)\cdot\frac{a}{b-c}=\frac{ab-b^2+c^2-ac}{bc}\cdot\frac{a}{b-c}\)

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

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

Do đó \(\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right)\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)=3+\frac{2a^3+2b^3+2c^3}{abc}=3+\frac{2\left(a^3+b^3+c^3\right)}{abc}=3+\frac{2.3abc}{abc}=9\)

=>đpcm

Sao phải phức tạp thế?

bài này có trong nâng cao phát triển toán 8 tập 1 nè!

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

Ta có : \(M.\frac{c}{a-b}=1+\frac{c}{a-b}\left(\frac{b-c}{a}+\frac{c-a}{b}\right)=+\frac{c}{a-b}\left(\frac{b^2-bc+ac-a^2}{ab}\right)\)

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

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

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

tên sai kìa,EKAWADA CONAN mà