Question

cho x+y+z=0 rút gọn:\(\dfrac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}\)

Answer 1

ta có: x+y+z=0

=>\(\left(x+y+z\right)^2=0=>x^2+y^2+z^2=-2\left(xy+yz+xz\right)\)

A=\(\dfrac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}=\dfrac{x^2+y^2+z^2}{y^2-2yz+z^2+z^2-2xz+x^2+x^2-2xy+y^2}=\dfrac{x^2+y^2+z^2}{2y^2+2z^2+2x^2-2\left(yz+xy+xz\right)}=\dfrac{x^2+y^2+z^2}{2y^2+2z^2+2x^2+x^2+y^2+z^2}=\dfrac{x^2+y^2+z^2}{3\left(x^2+y^2+z^2\right)}=\dfrac{1}{3}\)

vậy.......

chúc bạn học tốt ^ ^

Answer 2

Ta có :

\(\dfrac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}\)

⇔ \(\dfrac{x^2+y^2+z^2}{2\left(x^2+y^2+z^2\right)-2\left(xy+yz+xz\right)}\) (*)

Lại có :

\(x+y+z=0\)

⇔ \(\left(x+y+z\right)^2=0\)

⇔ \(x^2+y^2+z^2+2\left(xy+yz+xz\right)=0\)

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

Thay vào biểu thức (*) ta có :

\(\dfrac{-2\left(xy+yz+xz\right)}{-4\left(xy+yz+xz\right)-2\left(xy+yz+xz\right)}\)

= \(\dfrac{-2\left(xy+yz+xz\right)}{-6\left(xy=yz+xz\right)}\)

= \(\dfrac{1}{3}\)