Question

tìm x y thuộc R thoả mãn

\(\left(x+y+1\right)^2=3\left(x^2+y^2+1\right)\)

 

Answer 1

Ta có : \(\left(x+y+1\right)^2=3\left(x^2+y^2+1\right)=>x^2+y^2+1+2xy+2x+2y=3\left(x^2+y^2+1\right)\)

\(=>2x^2+2y^2+2-2xy-2x-2y=0\)

\(=>\left(x-y\right)^2+x^2-2x+1+y^2-2y+1=0\)

\(=>\left(x-y\right)^2+\left(x-1\right)^2+\left(y-1\right)^2=0\)

Mà \(\left(x-y\right)^2\ge0;\left(x-1\right)^2\ge0;\left(y-1\right)^2\ge0=>\left(x-y\right)^2+\left(x-1\right)^2+\left(y-1\right)^2\ge0\)

Dấu "=" xảy ra khi \(x-y=0,x-1=0,y-1=0=>x=y=1\)

Vậy x=y=1