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 ^^

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

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

abc  bnj k

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

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

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

Thay a=2c-b vào (1) và (2) ta được

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

Vậy a=b=c

Khi đó: \(P=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)

Nguồn: GV

Bảo Ngọc Đàm Bạn có chắc là ad đc tcdtsbn vs mọi a ; b ; c đôi một khác nhau ko ạ ?
nguyễn thị kim oanh              Trình bày bài kia là trg hợp 1 : a + b +  c ≠ 0 

Trường hợp 2 : a + b + c = 0 

~ Tự lm ~

Câu hỏi của Chu Hoàng THủy Tiên - Toán lớp 7 - Học toán với OnlineMath

Lời giải:

$M=\frac{-ab(a-b)}{(a-b)(b-c)(c-a)}+\frac{-bc(b-c)}{(a-b)(b-c)(c-a)}+\frac{-ca(c-a)}{(a-b)(b-c)(c-a)}$

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

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

xét a + b + c = 0 khi đó a + b = -c ; b + c = -a ; a + c = -b

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

xét a + b + c \(\ne\)0 . thì \(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)

\(\Rightarrow a+b=2c;b+c=2a\)\(\Rightarrow a-c=2\left(c-a\right)\)\(\Rightarrow a=c\)( loại vì a khác c )

Vậy A = -1

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

=> a= b =c 

=> P = (1+1) ( 1+1)(1+1) = 2.2.2 =8

cảm ơn