Theo tính chất dãy tỉ số bằng nhau ta có : a+b-c/c = b+c-a/a = c+a-b/b = a+b-c+b+c-a+c+a-b/a+b+c = a+b+c/a+b+c = 1

Ta có : a+b-c/c=1  => a+b-c=c  => a+b+c=3c   (1)

Ta có : b+c-a/a=1  => b+c-a=a  => a+b+c=3a   (2)

Ta có : c+a-b/b=1  => c+a-b=b  => a+b+c=3b   (3)

Từ (1);(2);(3)   => 3c=3a=3b  => a=b=c  => b/a=1 ; a/c=1 ; c/b=1

=> B= (1+b/a)(1+a/c)(1+c/b)  = (1+1)(1+1)(1+1) = 2.2.2 = 8

=8

8 8 cái địt mẹ mày

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

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

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

\(\Rightarrow a=b=c\)

\(\Rightarrow\frac{b}{a}=1;\frac{a}{c}=1;\frac{c}{b}=1\)

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

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

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

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

\(\Rightarrow a=b=c\)

\(\Rightarrow\frac{b}{a}=\frac{a}{c}=\frac{c}{b}=1\)

\(\Rightarrow B=2.2.2=8\)

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

 nếu a+b+c =0

=> a =0-b-c => a = -(b+c)

b     = 0-a-c => b = -(a+c)

c      = 0-a-b => c = -(a+b)

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

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

\(B=\frac{-c}{a}.\frac{-b}{c}.\frac{-a}{b}\)

\(B=-1\)

 nếu a+b+c khác 0

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

=> \(B=\left(1+\frac{b}{a}\right).\left(1+\frac{a}{c}\right).\left(1+\frac{c}{b}\right)\)

\(B=\left(1+1\right).\left(1+1\right).\left(1+1\right)\)

\(B=2.2.2\)

\(B=8\)

KL: B= -1 hoặc B=8

Chúc bn học tốt !!!!

Áp dụng tính chất của dãy tỉ số = nhau ta có:

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

Xét 2 trường hợp:

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

\(P=\frac{a+b}{a}.\frac{a+c}{c}.\frac{b+c}{b}\)

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

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

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

\(P=\frac{a+b}{a}.\frac{a+c}{c}.\frac{b+c}{b}=\frac{2c}{a}.\frac{2b}{c}.\frac{2a}{b}=8\)

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

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

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

a+b-c=c

2c=a+b

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

b+c-a=a

2a=b+c

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

c+a-b=b

=>c+a=2b

ta co \(P=\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)=\left(\frac{a+b}{a}\right)\left(\frac{a+c}{c}\right)\left(\frac{c+b}{b}\right)\)

=\(\frac{2c}{a}.\frac{2b}{c}.\frac{2a}{b}=2.2.2=8\)

Các cao nhân giúp với!!!!!!!!!! Thanks for all

Ta có:\(a+b+c\ne0\)vì nếu \(a+b+c=0\)thế vào giả thiết ta có:

\(\frac{a}{-a}+\frac{b}{-b}+\frac{c}{-c}=1\Leftrightarrow-3=1\)(vô lí)

Khi \(a+b+c\ne0\)ta có:

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

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

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

\(\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=0\)\(\Rightarrow P=0\)

Học tốt 

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

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

\(< =>P=a+b+c-a-b-c=0\)

Ta có:

(a+b-c)/c=(b+c-a)/a=(c+a-b)/b=(a+b-c+b+c-a+c+a-b)/(c+a+b)=0/(c+a+b)=0

=> a+b-c=0 =>a+b=c

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

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

=>B=(a+b)/a.(c+a)/c.(b+c)/b

      =c/a.b/c.a/b=1

TK!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Ta có:

(a+b-c)/c=(b+c-a)/a=(c+a-b)/b=(a+b-c+b+c-a+c+a-b)/(c+a+b)=0/(c+a+b)=0

=> a+b-c=0 =>a+b=c

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

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

=>B=(a+b)/a.(c+a)/c.(b+c)/b

      =c/a.b/c.a/b=1

Ta có : \(\left(1+\frac{b}{a}\right)\cdot\left(1+\frac{a}{c}\right)\cdot\left(1+\frac{c}{b}\right)\)=\(\frac{a+b}{a}\cdot\frac{a+c}{c}\cdot\frac{b+c}{b}\)(*)

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

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

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

ADTC dãy tỉ số = nhau ta có:

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

Suy ra : \(\frac{a+b}{c}\)=2  => a+b=2c   (1)

\(\frac{b+c}{a}\)=2  => b+c=2a   (2)

\(\frac{c+a}{b}\)=2  => c+a=2b   (3)

Thay (1);(2);(3) vào (*) ta có :

\(\frac{2c}{a}\cdot\frac{2b}{c}\cdot\frac{2a}{b}\)=\(\frac{\left(2\cdot2\cdot2\right)\cdot\left(a\cdot b\cdot c\right)}{a\cdot b\cdot c}\)=8