Question

Chmr nếu:

\(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}\) với \(x\ne y,yz\ne1,xz\ne1,x\ne0,y\ne0,z\ne0\)

thì: \(x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)

Answer 1

Áp dụng t/c dãy tỉ số bằng nhau có:

\(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}=\frac{x^2-yz-y^2+xz}{x-xyz-y\left(1-xz\right)}=\frac{\left(x-y\right)\left(x+y\right)+z\left(x-y\right)}{x-xyz-y+xyz}=\frac{\left(x-y\right)\left(x+y+z\right)}{x-y}=x+y+z\)

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

<=> \(\frac{x^2-yz}{x\left(1-yz\right)}-\frac{\left(x+y+z\right)x\left(1-yz\right)}{x\left(1-yz\right)}=0\)

<=> \(\frac{x^2-yz-\left(x^2+yx+zx\right)\left(1-yz\right)}{x\left(1-yz\right)}\)=0

<=> \(x^2-yz-x^2+x^2yz-xy+xy^2z-xz+xyz^2=0\)

<=> \(-yz-xy-xz+xyz\left(x+y+z\right)\)=0

<=> \(xyz\left(x+y+z\right)=yz+xy+xz\)

<=>\(x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)( chia cả hai vế cho xyz với x,y,z khác 0)