quá đơn giản 

cho 5 k giải cho

(mình trong đội tuyển toán đó nhe nên làm theo đi)

Từ: \(a^2\left(b+c\right)=b^2\left(a+c\right)\Leftrightarrow a^2b-ab^2+ca^2-cb^2=0\Leftrightarrow ab\left(a-b\right)+c\left(a-b\right)\left(a+b\right)=0.\)

\(\Leftrightarrow\left(a-b\right)\left(ab+bc+ac\right)=0\). Do \(a\ne b\Rightarrow ab+bc+ac=0\)(1)

Mặt khác, xét hiệu:

\(c^2\left(a+b\right)-a^2\left(b+c\right)=ac^2-a^2c+bc^2-a^2b=ac\left(c-a\right)+b\left(c-a\right)\left(c+a\right)=\)

\(=\left(c-a\right)\left(ac+bc+ab\right)=0\)

Do đó: \(H=c^2\left(a+b\right)=a^2\left(b+c\right)=2013.\)

đề ko sai đâu bạn

Từ đẳng thức \(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}\)

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

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

Nếu a + b + c = 0

=> a + b = - c;

b + c = - a;

c + a = - b

Khi đó M = \(\frac{-c}{2c}+\frac{-a}{a}+-\frac{b}{4b}=-\frac{1}{2}+\left(-1\right)+\left(-\frac{1}{4}\right)=-\frac{7}{4}=-1,75\)

Nếu a + b + c \(\ne\)0

=> b + c = c + a = a + b

=> a = b = c

Khi đó M = \(\frac{2c}{2c}+\frac{2a}{a}+\frac{2b}{4b}=1+2+\frac{1}{2}=\frac{7}{2}=3,5\)

Vậy nếu a + b + c = 0 thì M = -1,75

nếu a + b + c \(\ne\)0 thì M = 3,5

Ta có: \(\frac{ab + ac}{2} = \frac{bc + ba}{3} = \frac{ca + cb}{4}\)

=>\(\frac{a(b + c)}{2} = \frac{b(c + a)}{3} = \frac{c(a + b)}{4}\)

=>\(\frac{a(b + c)}{2abc} = \frac{b(c + a)}{3abc} = \frac{c(a + b)}{4abc}\)

=>\(\frac{b + c}{2bc} = \frac{c + a}{3ca} = \frac{a + b}{4ab}\)

=>\(\frac{b}{2bc} + \frac{c}{2bc} = \frac{c}{3ca} + \frac{a}{3ca} = \frac{a}{4ab} + \frac{b}{4ab}\)

=>\(\frac{1}{2c}+\frac{1}{2b}=\frac{1}{3a}+\frac{1}{3c}=\frac{1}{4b}+\frac{1}{4a}\)

=>\(12 \cdot \left(\frac{1}{2c} + \frac{1}{2b}\right) = 12 \cdot \left(\frac{1}{3a} + \frac{1}{3c}\right) = 12 \cdot \left(\frac{1}{4b} + \frac{1}{4a}\right)\)

=>\(\frac{6}{c}+\frac{6}{b}=\frac{4}{a}+\frac{4}{c}=\frac{3}{b}+\frac{3}{a}\quad(*)\)

\(\frac{6}{c} + \frac{6}{b} = \frac{4}{a} + \frac{4}{c}\)

=>\(\frac{6}{c}-\frac{4}{c}+\frac{6}{b}=\frac{4}{a}\)

=>\(\frac{2}{c}+\frac{6}{b}=\frac{4}{a}\quad(1)\)

Ta có: \(\frac{4}{a} + \frac{4}{c} = \frac{3}{b} + \frac{3}{a}\)

=>\(\frac{4}{a}-\frac{3}{a}+\frac{4}{c}=\frac{3}{b}\)

=>\(\frac{1}{a}+\frac{4}{c}=\frac{3}{b}\implies\frac{3}{b}-\frac{1}{a}=\frac{4}{c}\quad(2)\)

\(\frac{6}{c} + \frac{6}{b} = \frac{3}{b} + \frac{3}{a}\)

=>\(\frac{6}{b}-\frac{3}{b}+\frac{6}{c}=\frac{3}{a}\)

=>\(\frac{3}{b}+\frac{6}{c}=\frac{3}{a}\implies\frac{1}{b}+\frac{2}{c}=\frac{1}{a}\quad(3)\)

Từ (1),(3) suy ra \(\frac{2}{c} + \frac{6}{b} = 4 \left(\frac{1}{b} + \frac{2}{c}\right)\)

=>\(\frac{2}{c}+\frac{6}{b}=\frac{4}{b}+\frac{8}{c}\)

=>\(\frac{-6}{c}=\frac{-2}{b}\)

=>\(\frac{6}{c}=\frac{2}{b}\)

=>\(\frac{c}{6}=\frac{b}{2}\)

=>\(\frac{c}{15}=\frac{b}{5}\) (4)

=>\(\frac{1}{b}=\frac{3}{c}\)

=>\(\frac{1}{a} = \frac{3}{c} + \frac{2}{c} = \frac{5}{c}\)

=>\(\frac{a}{1}=\frac{c}{5}\)

=>\(\frac{a}{3}=\frac{c}{15}\) (5)

Từ (4),(5) suy ra \(\frac{a}{3}=\frac{b}{5}=\frac{c}{15}\) (ĐPCM)

\(\frac{ab+ac}{2}=\frac{bc+ab}{3}=\frac{ca+bc}{4}\)

( ta lần lược lấy - (1) + (2) + (3) = (1) - (2) + (3) = (1) + (2) - (3) được)

\(=\frac{2bc}{5}=\frac{2ca}{3}=\frac{2ab}{1}\)

Ta thấy rằng a,b,c không thể = 0 vì như vậy thì a + b + c \(\ne69\)

\(\Rightarrow\hept{\begin{cases}a=\frac{c}{5}\\b=\frac{c}{3}\end{cases}}\)

Thế vào: a + b + c = 69

\(\Leftrightarrow\frac{c}{5}+\frac{c}{3}+c=69\)

\(\Rightarrow c=45\)   

\(\Rightarrow\hept{\begin{cases}a=9\\b=15\end{cases}}\)  

Dùng tính chất dãy tỉ số bằng nhau mà làm

Ta có: \(\left(\frac{a+b+c}{b+c+d}\right)^3=\frac{\left(a+b+c\right)^3}{\left(b+c+d\right)^3}=\frac{a^3+b^3+c^3+2ab+2ac+2bc}{b^3+c^3+d^3+2bc+2bd+2cd}\)