Có : a/b+c = b/a+c = c/a+b => b+c/a = a+c/b = a+b/c

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

b+c/a = a+c/b = a+b/c = b+c+a+c+a+b/a+b+c = 2

=> P = 2+ 2 + 2  =6

k mk nha

SAI ROi

Vì  \(a+b+c=0\)  \(\Rightarrow\)  \(c=-a-b\)

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

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

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

Mặt khác, ta cũng có: từ \(a+b+c=0\), suy ra \(a^3+b^3+c^3=3abc\)

Vậy,  \(B=3+\frac{2\left(a^3+b^3+c^3\right)}{abc}=3+\frac{2.3abc}{abc}=3+6=9\)  (vì  \(a,b,c\ne0\)  nên \(abc\ne0\) )

cảm ơn bạn @@

Chứng minh bằng -1 à

thiếu đề à bạn ơi

đat x=\(\frac{a+b}{a-b}\) tu day suy ra \(x+1=\frac{2a}{a-b}\) \(x-1=\frac{2b}{a-b}\)

ttu \(y=\frac{b+c}{b-c}\Rightarrow y+1=\frac{2b}{b-c};y-1=\frac{2c}{b-c}\)

\(z=\frac{c+a}{c-a}\Rightarrow z+1=\frac{2c}{c-a};z-1=\frac{2a}{c-a}\)

ta sẻ có \(\left(x+1\right)\left(y+1\right)\left(z+1\right)=\left(x-1\right)\left(y-1\right)\left(z-1\right)\) (bn chịu khó cm nhé_

khai triên ra ta sẽ có \(xy+yz+xz=-1\) suy ra dpcm

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

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

abc  bnj k

Lời giải:

Ta có:

$\frac{a+b}{a-b}.\frac{b+c}{b-c}+\frac{a+b}{a-b}.\frac{c+a}{c-a}+\frac{b+c}{b-c}.\frac{c+a}{c-a}$

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

$=\frac{[b^2+(ab+bc+ac)](c-a)+[a^2+(ab+bc+ac)](b-c)+[c^2+(ab+bc+ac)](a-b)}{(a-b)(b-c)(c-a)}$

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

$=\frac{b^2(c-a)+a^2(b-c)+c^2(a-b)}{(a-b)(b-c)(c-a)}$

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

Ta có đpcm.

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

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

tick nha công mk đánh máy

Ta có:

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

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

\(\Rightarrow P=\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=3.2=6\)

bài này có 2 trường hợp nhé =))

\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}\Rightarrow1+\frac{a}{b+c}=1+\frac{b}{a+c}=1+\frac{c}{a+b}\)

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

\(TH1:a+b+c=0\)

\(\Rightarrow\hept{\begin{cases}b+c=-a\\a+c=-b\\a+b=-c\end{cases}\Rightarrow P=\frac{-a}{a}+\frac{-b}{b}+\frac{-c}{c}=-3}\)

\(TH2:a+b+c\ne0\)

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

\(\Rightarrow P=\frac{a+a}{a}+\frac{b+b}{b}+\frac{c+c}{c}=2.3=6\)

Vậy P=-3 hay P=6

Ko hiểu ???