Question

[Fuiki Fuiko]

Cho x,y,z thỏa mãn đk x/(y+z)+y/(x+z)+z/(x+y)=1

Tính giá trị của S=x^2/(y+z)+y^2/(x+z)+z^2/(x+y)

Answer 1

[ST]

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

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

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

<=>\(S=\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}=0\)

Answer 2

[đoàn minh Hải]

 x/(y+z)+y/(x+z)+z/(x+y)=1

=>\(\frac{x^2}{\left(y+z\right)^2}\)+\(\frac{y^2}{\left(x+z\right)^2}\)+\(\frac{z^2}{\left(x+y\right)^2}\)+2(\(\frac{xy}{\left(y+z\right)\cdot\left(x+z\right)}\)+\(\frac{yz}{\left(x+z\right)\left(x+y\right)}\)+\(\frac{zx}{\left(z+y\right)\cdot\left(x+y\right)}\))=1