Question

Tìm x,y,z biết: \(x^2+y^2+z^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}=6\)

Answer 1

\(\left(x^2+\dfrac{1}{x^2}\right)+\left(y^2+\dfrac{1}{y^2}\right)+\left(z^2+\dfrac{1}{z^2}\right)=6\)

=> VT≥\(2.\sqrt{x^2.\dfrac{1}{x^2}}+2.\sqrt{y^2.\dfrac{1}{y^2}}+2.\sqrt{z^2.\dfrac{1}{z^2}}\)

= 2+2+2=6

Dau bang xay ra khi: \(\left\{{}\begin{matrix}x^2=\dfrac{1}{x^2}\\y^2=\dfrac{1}{y^2}\\z^2=\dfrac{1}{z^2}\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=\pm1\\y=\pm1\\z=\pm1\end{matrix}\right.\)

Answer 2

\(x^2+y^2+z^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}=6\) =>\(\left(x^2+2+\dfrac{1}{x^2}\right)+\left(y^2+2+\dfrac{1}{y^2}\right)+\left(z^2+2+\dfrac{1}{z^2}\right)=0\) =>\(\left(x+\dfrac{1}{x}\right)^2+\left(y+\dfrac{1}{y}\right)^2+\left(z+\dfrac{1}{z}\right)^2=0\) =>\(\left(\dfrac{x^2+1}{x}\right)^2+\left(\dfrac{y^{^2}+1}{y}\right)^2+\left(\dfrac{z^2+1}{z}\right)^2=0\) x^2+1=0=>x^2=-1 y^2+1=0=>y^2=-1 z^2+1=0=>z^2=-1 =>x,y,z k có gt thỏa mãn tớ k chắc cho lắm ==