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Đặt \(A=\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}=2009,B=\frac{y^2}{x+y}+\frac{z^2}{y+z}+\frac{z^2}{x+z}\)
\(=>A-B=\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{x^2}{z+x}-\frac{y^2}{x+y}-\frac{z^2}{y+z}+\frac{x^2}{z+x}\)
\(=>2009-B=\frac{x^2-y^2}{x+y}+\frac{y^2-z^2}{y-z}+\frac{z^2-x^2}{z-x}\)
\(=>2009-B=\frac{\left(x-y\right).\left(x+y\right)}{x+y}+\frac{\left(y-z\right).\left(y+z\right)}{y+z}+\frac{\left(z-x\right).\left(z+x\right)}{z+x}\)
=>2009-B=x-y+y-x+z-x
=>2009-B=(x-x)+(y-y)+(z-z)
=>2009-B=0+0+0
=>2009-B=0
=>B=2009
Vậy \(\frac{x^2}{x+y}+\frac{z^2}{y+z}+\frac{x^2}{z+x}=2009\)
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?
\(\Rightarrow\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right).\left(x+y+z\right)=x+y+z\)
\(\Rightarrow\frac{x^2+x\left(z+x\right)}{y+z}+\frac{y^2+y\left(x+z\right)}{x+z}+\frac{z^2+z\left(x+y\right)}{x+y}=x+y+z\)
\(\Rightarrow\frac{x^2}{y+z}+x+\frac{y^2}{x+z}+y+\frac{z^2}{x+y}+z=x+y+z\)
\(\Rightarrow\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}=0\)
\(\Rightarrow\left(\frac{x}{x+y}+\frac{y}{z+x}+\frac{z}{x+y}\right)\cdot\left(x+y+z\right)=x+y+z\)
\(\Rightarrow\frac{x^2}{y+z}+\frac{xy}{y+z}+\frac{xz}{y+z}+\frac{y^2}{z+x}+\frac{xy}{z+x}+\frac{yz}{z+x}+\frac{z^2}{x+y}+\frac{xz}{x+y}+\frac{yz}{x+y}=x+y+z\)
Rồi bạn cộng 2 phân thức 2,3 5,6 8,9 lại thì được
\(\Rightarrow\frac{x^2}{y+z}+x+\frac{y^2}{z+x}+y+\frac{z^2}{x+y}+z=x+y+z\)
\(\Rightarrow\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}=0\)
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 \(\left(\frac{y^2}{x+y}+\frac{z^2}{y+z}+\frac{x^2}{z+x}\right)=2009\)