ta có a+b+c=0       =>     a=-b-c,         b=-a-c,            c=-a-b

thay vào A ta được 

 A=(1-(b+c)/b)(1-(a+c)/c)(1-(a+b)/a)

   =(1-1-c/b)(1-1-a/c)(1-1-b/a)

   =(-c/b)(-a/c)(-b/a)

   =(-abc)/abc

    =-1

bạn Nguyễn Thị Lan Hương làm đúng rồi, mk lm cách khác nhé:

           BÀI LÀM

          \(a+b+c=0\)

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

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

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

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

1. Ta có : \(\left(\frac{1}{a}-\frac{1}{b}\right)^2\ge0\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{2}{ab}\)

Tương tự :  \(\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{2}{bc}\); \(\frac{1}{a^2}+\frac{1}{c^2}\ge\frac{2}{ac}\)

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\). Dấu " = " xảy ra \(\Leftrightarrow\)a = b = c

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=9\)

\(9\le3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge3\)

Dấu " = " xảy ra \(\Leftrightarrow\)a = b = c = 1

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=7\)\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=49\)

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\frac{a+b+c}{abc}=49\)

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=49\)

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

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

\(\frac{\left(c-b\right)\left(a-c\right)\left(a-b\right)}{abc}\)

Ta thấy c -b \(\ge\)0 ; a - c \(\le\)0 ; a - b \(\le\)0 nên ( c - b ) ( a - c ) ( a - b )\(\ge\)0

Mà abc > 0 nên A \(\ge\)0 => ....

Ta có:

Dấu “=” xảy ra khi và chỉ khi

Vậy số bộ a,b,c thỏa mãn điều kiện đã cho là 1.

Chọn B.

mk nhầm 1+c/a

Ta có : \(a+b+c\Rightarrow\hept{\begin{cases}a+b=-c\\b+c=-a\\a+c=-b\end{cases}\left(\cdot\right)}\)

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

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

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

\(=-1.-1.-1\)

\(=-1\)

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

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

<=>c(a+b)(a+b+c)=-ab(a+b)

<=>(a+b)(ac+bc+c²)+ab(a+b)=0

<=>(a+b)(ac+bc+ab+c²)=0

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

       a+b=0

<=> b+c=o

       c+a=0
 

Từ 1a+1b+1c=0⇒ab+bc+ac=01a+1b+1c=0⇒ab+bc+ac=0

Khi đó:

(√a+c+√b+c)2=a+c+b+c+2√(a+c)(b+c)(a+c+b+c)2=a+c+b+c+2(a+c)(b+c)

=a+b+2c+2√ab+ac+bc+c2=a+b+2c+2√c2=a+b+2c+2ab+ac+bc+c2=a+b+2c+2c2

=a+b+2c+2|c|=a+b+2c+2|c|

Vì a,ba,b dương nên −1c=1a+1b>0⇒c<0⇒2|c|=−2c−1c=1a+1b>0⇒c<0⇒2|c|=−2c

Do đó:

(√a+c+√b+c)2=a+b+2c+2|c|=a+b+2c+(−2c)=a+b(a+c+b+c)2=a+b+2c+2|c|=a+b+2c+(−2c)=a+b

⇒√a+c+√b+c=√a+b

Từ 1a+1b+1c=0⇒ab+bc+ac=01a+1b+1c=0⇒ab+bc+ac=0

Khi đó:

(√a+c+√b+c)2=a+c+b+c+2√(a+c)(b+c)(a+c+b+c)2=a+c+b+c+2(a+c)(b+c)

=a+b+2c+2√ab+ac+bc+c2=a+b+2c+2√c2=a+b+2c+2ab+ac+bc+c2=a+b+2c+2c2

=a+b+2c+2|c|=a+b+2c+2|c|

Vì a,ba,b dương nên −1c=1a+1b>0⇒c<0⇒2|c|=−2c−1c=1a+1b>0⇒c<0⇒2|c|=−2c

Do đó:

(√a+c+√b+c)2=a+b+2c+2|c|=a+b+2c+(−2c)=a+b(a+c+b+c)2=a+b+2c+2|c|=a+b+2c+(−2c)=a+b

⇒√a+c+√b+c=√a+b