\(a+b+c\ne0\) biết a = 2003

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

\(\frac{a}{b}=\frac{c}{a}\Rightarrow bc=a^2=2003^2\)

\(\Rightarrow b=2003;c=2003\\\)

Vậy b = 2003;c = 2003

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

\(\begin{cases}=>a=b\\=>b=c\\=>c=a\end{cases}=>a=b=c}\)

\(b=2003;c=2003\)

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

\(=>a=b\)

\(=>b=c\)

\(=>c=a\)

\(< =>a=b=c\)

\(=>b=2003;c=2003\)

vậy :\(b=2003;c=2003\)

theo bài ra ta có:

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

áp dụng t/c dãy tỉ số bằng nhau ta có:

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

=> a = b

b = c

c = a

=> a = b =c

mà a = 2003 => b = c = 2003

vậy b = 2003, c = 2003

\(P=\frac{ab+c}{\left(a+b\right)^2}.\frac{bc+a}{\left(b+c\right)^2}.\frac{ca+b}{\left(c+a\right)^2}\)

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

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

\(\)Ta có: \(a+b+c=0 \Rightarrow b+c=-a \Rightarrow (b+c)^2=(-a)^2 \Leftrightarrow b^2+c^2+2bc=a^2 \Leftrightarrow a^2-b^2-c^2=2bc\)

Tương tự: \(b^2-c^2-a^2=2ca;c^2-a^2-b^2=2ab\)

\(P=...=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ca}+\dfrac{c^2}{2bc}=\dfrac{a^3+b^3+c^3}{2abc}=\dfrac{3abc}{2abc}=\dfrac{3}{2}\)

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Bổ đề \(a+b+c=0 \Leftrightarrow a^3+b^3+c^3\)

Ở đây ta c/m chiều thuận:
Với \(a+b+c=0 \Leftrightarrow a+b=-c \Rightarrow (a+b)^3=(-c)^3 \Leftrightarrow a^3+b^3+3ab(a+b)=-c^3 \Leftrightarrow a^3+b^3+c^3=3abc(QED)\)

áp dụng tính chất dẫy tỉ số = nhau ta được 

b+c+d/a=c+d+a/b=a+b+d/c=a+b+c/d= b+c+d+c+d+a+a+b+d+a+b+c / a+b+c+d = 3 

do b+c+d/a=c+d+a/b=a+b+d/c=a+b+c/d = k 

suy ra k =3 .đơn giản vậy thôi

k = 3  có đúng ko bạn 

Ta có: \(\frac{a+b}{b+c}=\frac{c+d}{d+a}.\)

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

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

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

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

Nếu \(a+b+c+d\ne0.\)

\(\Rightarrow c+d=d+a\)

\(\Rightarrow c=a\left(đpcm1\right).\)

Nếu \(a+b+c+d=0\) thì hợp với đề.

\(\Rightarrow a+b+c+d=0\left(đpcm2\right).\)

Chúc bạn học tốt!

Từ \(a+b+c=0\) bạn tự chứng minh \(a^3+b^3+c^3=3abc\)

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

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

                   \(=1+\frac{2c^2}{ab}=1+\frac{2c^3}{abc}\)

Tương tự, ta có: \(A=3+\frac{2\left(a^3+b^3+c^3\right)}{abc}=3+\frac{2.3abc}{abc}=3+6=9\)