ddap an la bang -1 

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

Vì \(a+b+c=0\)

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

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

Ta có: \(a+b+c=0\)

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

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

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

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

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

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

\(\Rightarrow A=-1\)

~~k cho mik nha~~

cm \(P\ge\frac{3}{4}\)nhé mn

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

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

=> a=b=c

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

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

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

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

dễ ợt mày ngu thế

moi nguoi oi hom truoc minh hoc tap hop cac so TN do thi co cua minh day nhu sau 

vd: A={xeN/3<x<9}

thi minh liet ke ra la A=4,5,6,7,8 nhung sua bai lai ko dung 

co sua nhu vay A=3,4,5,6,7,8

ko biet hay sai mong ae giup minh

Áp dụng BĐT Cô-si \(ab\le\frac{\left(a+b\right)}{4}^2\)

=> \(\left(2a+b\right)\left(2c+b\right)\le\frac{4\left(a+b+c\right)^2}{4}=\left(a+b+c\right)^2\)

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

Mấy cái kia làm tương tự cậu nhé 

Dấu "=" xảy ra khi và chỉ khi a=b=c=1