Cho `xy+yz+zx=0` Tính `A=(yz)/(x^2+2yz)+(xz)/(y^2+2xz)+(xy)/(z^2+2xy)`

Đáp án + Giải thích các bước giải:

Ta có: `xy+yz+zx=0`

`=>{(xy=-yz-zx),(yz=-xy-zx),(zx=-xy-yz):}`

Ta có: `A=(yz)/(x^2+2yz)+(xz)/(y^2+2xz)+(xy)/(z^2+2xy)`

`A=(yz)/(x^2+yz+yz)+(xz)/(y^2+xz+xz)+(xy)/(z^2+xy+xy)`

`A=(yz)/(x^2-xy-zx+yz)+(xz)/(y^2-yz-xy+xz)+(xy)/(z^2-yz-zx+xy)`

`A=(yz)/(x(x-y)-z(x-y))+(xz)/(y(y-z)-x(y-z))+(xy)/(z(z-y)-x(z-y))`

`A=(yz)/((x-y)(x-z))-(xz)/((x-y)(y-z))+(xy)/((y-z)(x-z))`

`A=(yz(y-z))/((x-y)(y-z)(x-z))-(xz(x-z))/((x-y)(y-z)(x-z))+(xy(x-y))/((x-y)(y-z)(x-z))`

`A=(y^2z-yz^2-x^2z+xz^2+x^2y-xy^2)/((x-y)(y-z)(x-z))`

`A=(-z(x^2-y^2)-z^2(y-x)+xy(x-y))/((x-y)(y-z)(x-z))`

`A=(-z(x+y)(x-y)+z^2(x-y)+xy(x-y))/((x-y)(y-z)(x-z))`

`A=((x-y)(-zx-yz+z^2+xy))/((x-y)(y-z)(x-z))`

`A=((x-y)[x(y-z)-z(y-z)])/((x-y)(y-z)(x-z))`

`A=((x-y)(y-z)(x-z))/((x-y)(y-z)(x-z))`

`A=1`

Vậy `A=1`