Question

(2x-5)²+(3y+4)⁴+(2z-1)⁸\(\le\)0.Tìm x,y,z

Giúp mik vs!~~~

Answer 1

(2x-5)²+(3y+4)⁴+(2z-1)⁸ \(\le\) 0 (1)

Có: (2x-5)²\(\ge0\forall x\); (3y+4)⁴\(\ge0\forall y\); (2z-1)⁸\(\ge0\forall z\)

\(\Rightarrow\) (2x-5)²+(3y+4)⁴+(2z-1)⁸\(\ge0\forall x,y,z\) (2)

Từ (1); (2) \(\Rightarrow\left\{{}\begin{matrix}\left(2x-5\right)^2=0\\\left(3y+4\right)^4=0\\\left(3z-1\right)^8=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2x-5=0\\3y+4=0\\2z-1=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2x=5\\3y=-4\\2z=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{5}{2}\\y=\frac{-4}{3}\\z=\frac{1}{2}\end{matrix}\right.\)

Vậy .....

Answer 2

25 taoo viet 2=76 ra 26569