Xét hiệu :

\(\left(\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}\right)-\left(\frac{y^2}{x+y}+\frac{z^2}{y+z}+\frac{x^2}{z+x}\right)\)

\(=\frac{x^2-y^2}{x+y}+\frac{y^2-z^2}{y+z}+\frac{z^2-x^2}{z+x}\)

\(=\frac{\left(x+y\right)\left(x-y\right)}{x+y}+\frac{\left(y+z\right)\left(y-z\right)}{y+z}+\frac{\left(z+x\right)\left(z-x\right)}{z+x}\)

\(=x-y+y-z+z-x=0\)

Vậy \(\left(\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}\right)=\left(\frac{y^2}{x+y}+\frac{z^2}{y+z}+\frac{x^2}{z+x}\right)\)

hay \(\left(\frac{y^2}{x+y}+\frac{z^2}{y+z}+\frac{x^2}{z+x}\right)=2009\)

a) \(A=\frac{2}{x-y}+\frac{2}{y-z}+\frac{2}{z-x}+\frac{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)

         \(=\frac{2\left(y-z\right)\left(z-x\right)+2\left(x-y\right)\left(z-x\right)+2\left(x-y\right)\left(y-z\right)+\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)

           \(=\frac{\left[\left(x-y\right)+\left(y-z\right)+\left(z-x\right)\right]^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=\frac{\left(x-y+y-z+z-x\right)^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=0\)

Áp dụng: \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)

b)Ta có: \(\frac{x^2}{y+z}+x=\frac{x^2+x\left(y+z\right)}{y+z}=\frac{x^2+xy+xz}{y+z}=\frac{x\left(x+y+z\right)}{y+z}\)

    Tương tự:   \(\frac{y^2}{x+z}+y=\frac{y^2+xy+zy}{x+z}=\frac{y\left(x+y+z\right)}{x+z}\)

                \(\frac{z^2}{x+y}+z=\frac{z^2+xz+zy}{x+y}=\frac{z\left(x+y+z\right)}{x+y}\)

Suy ra: \(A+\left(x+y+z\right)\)

\(=\frac{x\left(x+y+z\right)}{y+z}+\frac{y\left(x+y+z\right)}{z+x}+\frac{z\left(x+y+z\right)}{x+y}+\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}+1\right)\)

  \(=2.\left(x+y+z\right)\)

Nên \(A=2.\left(x+y+z\right)-\left(x+y+z\right)=x+y+z\)

Mình có sai chỗ nào không nhỉ?

Xét hiệu của hai phân thức sau:

\(\left(\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}\right)-\left(\frac{y^2}{x+y}+\frac{z^2}{y+z}+\frac{x^2}{z+x}\right)=\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}-\frac{y^2}{x+y}-\frac{z^2}{y+z}-\frac{x^2}{z+x}\)

\(=\left(\frac{x^2}{x+y}-\frac{y^2}{x+y}\right)+\left(\frac{y^2}{y+z}-\frac{z^2}{y+z}\right)+\left(\frac{z^2}{z+x}-\frac{x^2}{z+x}\right)=x-y+y-z+z-x=0\)

Vì hiệu của chúng bằng  \(0\)  nên số bị trừ sẽ bằng số trừ, tức là:

\(\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}=\frac{y^2}{x+y}+\frac{z^2}{y+z}+\frac{x^2}{z+x}\)

Mà  \(\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}=2015\)  (theo giả thiết)

Vậy,  \(\frac{y^2}{x+y}+\frac{z^2}{y+z}+\frac{x^2}{z+x}=2015\)

Vì hiệu của chúng bằng 0 nên số bị trừ sẽ bằng số trừ ,tức là:

x^2/x+y+y^2/y+z+z^2/z+x=y^2/x+y+z^2/y+z+x^2/z+x

Mà x^2/x+y+y^2/y+z+z^2/z+x=2015(giả thiết)

Vậy y^2/x+y+z^2/y+z+x^2/z+x=2015

i don't know

\(\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)\left(x+y+z_{ }\right)=x+y+z\)+z

\(\frac{x^2}{y+z}+x+\frac{y^2}{x+z}+y+\frac{z^2}{x+y}+z=x+y+z\)

suy ra S=0

\(\Rightarrow\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right).\left(x+y+z\right)=x+y+z\)

\(\Rightarrow\frac{x^2+x\left(z+x\right)}{y+z}+\frac{y^2+y\left(x+z\right)}{x+z}+\frac{z^2+z\left(x+y\right)}{x+y}=x+y+z\)

\(\Rightarrow\frac{x^2}{y+z}+x+\frac{y^2}{x+z}+y+\frac{z^2}{x+y}+z=x+y+z\)

\(\Rightarrow\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}=0\)

dùng tính chất của dãy tỉ số bằng nhau