Question

[Linh Linh]

Tìm x;y;z \(thuộc\) Z sao cho

x²=y-1

y²=z-1

z²=x-1

Các bn ơi mik cần gấp nha!!!!!

Answer 1

[Xuân Tuấn Trịnh]

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

Answer 2

[Nguyễn Chí Mãi]

xin lổi nhe