\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{b+c+a+c+a+b}=\frac{a+b+c}{2a+2b+2c}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)\(\Rightarrow b+c=2a\)

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

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

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

\(D=\frac{2a}{a}=\frac{2b}{b}=\frac{2c}{c}\)

\(D=2+2+2\)

\(D=6\)

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

=>b+c=2a

=>a+c=2b

=>a+b=2c

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

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

\(D=2+2+2\)

D=6

Vậy D=6

 ^...^  ^_^

⇒a+c=2b⇒a+c=2b

⇒a+b=2c⇒a+b=2c

D=b+c/a+a+c/b+a+b/c

D=2a/a=2b/b=2c/c

D=2+2+2

D=6

Vì \(a,b,c\ne0\) nên:

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

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

\(\Rightarrow D=\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=\frac{2a}{a}+\frac{2b}{b}+\frac{2c}{c}=2+2+2=6\)

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

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

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

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

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

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

\(D=2+2+2\)

\(D=6\)

Vậy \(D=6\)

Xin lỗi 2 bạn kết bạn bằng 3

Cho a, b, c khác 0 thoả mãn a+b+c=0. Tính $A=\left(1+\frac{a}{b}\right)+\left(1+\frac{b}{c}\right)+\left(1+\frac{c}{a}\right)$A=(1+ab )+(1+bc )+(1+ca )

Cho a, b, c khác 0 thoả mãn a+b+c=0. Tính $A=\left(1+\frac{a}{b}\right)+\left(1+\frac{b}{c}\right)+\left(1+\frac{c}{a}\right)$A=(1+ab )+(1+bc )+(1+ca )

Khó quá do anh thien

\(A=3+\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\)

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

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

Thiếu đề nha bạn 

Cho a, b, c khác 0 thoả mãn a+b+c=0. Tính 

Bạn tham khảo câu hỏi tương tự. 

Câu hỏi của Đào Thị Lan Nhi - Toán lớp 7 - Học trực tuyến OLM

A=\(\frac{a^2}{bc}\)+\(\frac{b^2}{ac}\)+\(\frac{c^2}{ab}\)=\(\frac{a^3}{abc}\)+\(\frac{b^3}{abc}\)+\(\frac{c^3}{abc}\)=\(\frac{a^3+b^3+c^3}{abc}\)

Mà a^3+b^3+c^3=3abc ( Tự chứng minh )

\(\Rightarrow\)A= \(\frac{3abc}{abc}\)= 3

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vì a+b+c=0 => a+b= -c; b+c=-a; c+a=-b

(1+a/b)(1+b/c)(1+c/a)

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

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

=-1

Thay a = -2 ; b = 1 ; c = 1 ( vì -2 + 1 + 1 = 0 )

Ta có : \(A=\left(1+\frac{-2}{1}\right)\left(1+\frac{1}{1}\right)\left(1+\frac{1}{-2}\right)\)

           \(A=-1.2..\frac{1}{2}\)

           \(A=-1\)

\(1\)

nobita kun tính mò giá trị của a,b và c à