\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\frac{1}{a}+\frac{1}{b}=\frac{-1}{c}\Rightarrow\frac{a+b}{ab}=\frac{-1}{c}\)

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

Tương tự : \(b+c=\frac{-bc}{a};a+c=\frac{-ac}{b}\)

thay vào A,ta được :

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

nhầm đoạn cuối : \(A=\frac{-a^2b^2c^2}{a^2b^2c^2}=-1\)

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

Vì \(a+b+c=0\)

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

\(\Rightarrow A=\frac{-c}{b}.\left(\frac{-a}{c}\right).\left(\frac{-b}{a}\right)\)

\(\Rightarrow A=-1\)

toán nâng cao ak bn

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

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

=> a=b=c

A=(1+1)(1+1)(1+1) = 2.2.2 =8

\(\sqrt{\frac{\left(a+bc\right)\left(b+ac\right)}{c+ab}}=\sqrt{\frac{\left(a^2+ab+ac+bc\right)\left(b^2+bc+ba+ac\right)}{c^2+ca+cb+ab}}=\sqrt{\frac{\left(a+b\right)\left(a+c\right)\left(b+a\right)\left(b+c\right)}{\left(c+a\right)\left(c+b\right)}}=a+b\left(a,b,c>0;a+b+c=1\right)\)

Bạn làm tương tự nha

\(\Rightarrow P=a+b+c+a+b+c=2\left(a+b+c\right)=2\)

help me

=>\(\frac{a-b+c}{2b}+1=\frac{c-a+b}{2a}+1=\frac{a-c+b}{2c}+1\)

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

*TH1: nếu a+b+c=0 => a+b=-c; b+c=-a; c+a=-b

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

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

*TH2: Nếu a+b+c khác 0: thì a=b=c

Khi đó P=2.2.2=8

Vậy P= -1 hoặc 8

TH1 a+b+c khác 0 

Aps dụng tính......a−b+c2b =c−a+b2a =a−c+b2c =1/2 (tự tính )

* với a-b+c/2b=1/2 suy ra 2a-2b+2c=2b suy ra 2a+2c=4b suy ra 2(a+c)=2.2.b suy ra a+c =2b

tương tự với từng t/h một  thì ta được c+b=2a;a+b=2c

suy raP=(1+cb )(1+ba )(1+ac ) =(b+c/b)(a+b/a)(c+a/c)=2a/b.2c/a.2a/c=8

TH2a+b+c =0 tự cm

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

\(\Leftrightarrow\frac{ab+bc+ac}{abc}=\frac{1}{a+b+c}\Leftrightarrow\left(a+b+c\right)\left(ab+bc+ac\right)-abc=0\)

\(\Leftrightarrow a^2b+a^2c+b^2a+b^2c+abc+abc+bc^2+ac^2=0\)

\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\Leftrightarrow...\)

\(P=0\)