Từ đề bài:
=>x2+y2+z2=x+y+z-3
<=>x2-x+\(\dfrac{1}{4}+y^2-y+\dfrac{1}{4}+z^2-z+\dfrac{1}{4}+\dfrac{9}{4}\)=0
<=>\(\left(x-\dfrac{1}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2+\left(z-\dfrac{1}{2}\right)^2+\dfrac{9}{4}=0\)(1)
Do \(\left(x-\dfrac{1}{2}\right)^2\)\(\ge0\forall x\in R\)
\(\left(y-\dfrac{1}{2}\right)^2\)\(\ge0\forall y\in R\)
\(\left(z-\dfrac{1}{2}\right)^2\)\(\ge0\forall z\in R\)
=>\(\left(x-\dfrac{1}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2+\left(z-\dfrac{1}{2}\right)^2\)\(\ge0\forall x;y;z\in R\)
\(\left(x-\dfrac{1}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2+\left(z-\dfrac{1}{2}\right)^2+\dfrac{9}{4}\)\(\ge\dfrac{9}{4}>0\forall x;y;z\in R\)
=>(1) vô nghiệm
Vậy không tồn tại x,y,z thỏa mãn đề bài
