Question

toan 8 cho (x+y)(x+z)+(y+z)(y+x)=2(z+x)(z+y) chung minh z^2=x^2+y^2/2

Answer 1

Ta có :

\(\left(x+y\right)\left(x+z\right)+\left(y+z\right)\left(y+x\right)\)

\(=x^2+xz+xy+yz+y^2+xy+zy+xz\)

\(=x^2+y^2+2\left(xz+xy+yz\right)\)

\(2\left(x+z\right)\left(z+y\right)\)

\(=2\left(xz+z^2+xy+zy\right)\)

\(=2z^2+2\left(xz+xy+yz\right)\)

\(\Rightarrow x^2+y^2+2\left(xz+xy+yz\right)=2z^2+2\left(xz+xy+yz\right)\)

\(\Rightarrow x^2+y^2=2z^2\)

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

Vây ...