\(\left(3x-5\right)^{2018}+\left(y^2-1\right)^{2006}+\left(x-z\right)^{2100}=0\)

ta có \(\left\{{}\begin{matrix}\left(x-z\right)^{2100}\ge0\\\left(y^2-1\right)^{2006}\ge0\\\left(3x-5\right)^{2018}\ge0\end{matrix}\right.\)

dấu = xảy ra khi \(\left\{{}\begin{matrix}3x-5=0\\y^2-1=0\\z-x=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}\\z=x\\\left[{}\begin{matrix}y=1\\y=-1\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\dfrac{5}{3}\\y=1\\z=\dfrac{5}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{5}{3}\\y=-1\\z=\dfrac{5}{3}\end{matrix}\right.\end{matrix}\right.\)

vậy.................

\(\left(3x-5\right)^{2006}+\left(y^2-1\right)^{2008}+\left(x-z\right)^{2100}=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(3x-5\right)^{2006}=0\\\left(y^2-1\right)^{2008}=0\\\left(x-z\right)^{2100}=0\end{matrix}\right.\Leftrightarrow x=z=\dfrac{5}{3}\)

\(\Rightarrow\left[{}\begin{matrix}y=1\\y=-1\end{matrix}\right.\)

Từ đề suy ra :

\(\left\{{}\begin{matrix}\left(3x-5\right)^{2006}=0\\\left(y^2-1\right)^{2008}=0\\\left(x-z\right)^{2100}=0\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}3x-5=0\\y^2-1=0\\x-z=0\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=z=\dfrac{5}{3}\\y=\pm1\end{matrix}\right.\)

(3x-5)²⁰⁰⁶ + (y²-1)²⁰⁰⁸ + (x-z)²¹⁰⁰ = 0

Vì (3x-5)²⁰⁰⁶, (y²-1)²⁰⁰⁸ , (x-z)²¹⁰⁰ > hoặc =0 ( với mọi x, y, z)

=>(3x-5)²⁰⁰⁶ =0 hoặc (y²-1)²⁰⁰⁸ =0 hoặc (x-z)²¹⁰⁰ =0

=>3x-5 =0 =>y²-1 =0 =>x-z =0

=>3x =5 =>y² =1 => x = z = 5/3

=> x =5/3 =>y=1 hoặc y=-1

Vậy (x;y;z)=(5/3; 1; 5/3) , (5/3; -1; 5/3)

đề sai 1 tí

Trên kia là (y²-1)⁶²⁰⁰⁸ đúng ko bạn?

nhấn vào đúng 0 sẽ ra

x=11;y=3;z=0

thiếu đề bnj ơi

https://olm.vn/hoi-dap/question/925051.html

bn vào link này có bài bạn caamnf đo. mà đề (3x-5)²⁰⁰⁶+(y²-1)²⁰⁰⁸+(x-z)²¹⁰⁰=0

đề bài là: (3x-5)²⁰⁰⁶+(y²-1)²⁰⁰⁸+(x-z)²¹⁰⁰=0 nha