Question

[nguyen thu phuong]

B1: 

Answer 1

[Edogawa Conan]

1. 

A = \(\frac{x^2-10y^2+10z^2}{y+z}+\frac{y^2-10z^2+10x^2}{z+x}+\frac{z^2-10x^2+10y^2}{x+y}\)

A = \(\frac{x^2}{y+z}-10\frac{\left(y-z\right)\left(y+z\right)}{y+z}+\frac{y^2}{x+z}-10\frac{\left(z-x\right)\left(z+x\right)}{z+x}+\frac{z^2}{x+y}-10\frac{\left(x-y\right)\left(x+y\right)}{x+y}\)

A = \(\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}-10y+10z-10z+10x-10x+10y\)

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

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

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

A = \(\left(x+y+z\right)-\left(x+y+z\right)=0\)