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

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

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

Từ đề suy ra :

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

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

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

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 

Ta có: \(\left(3x-5\right)^{2006}\ge0\)với mọi x

           \(\left(y^2-1\right)^{2008}\ge0\)với mọi y

           \(\left(x-z\right)^{2100}\ge0\) với mọi x,z

\(\Rightarrow\)\(\left(3x-5\right)^{2006}+\left(y^2-1\right)^{2008}+\left(x-z\right)^{2100}\ge0\)với mọi x

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

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

Xét:

\(\left(3x-5\right)^{2006}=0\hept{\begin{cases}3x-5=0\\3x=5\\x=\frac{5}{3}\end{cases}}\)

Xét:

\(\left(y^2-1\right)^{2008}=0\hept{\begin{cases}y^2-1=0\\y^2=1\\y=1hoac-1\end{cases}}\)

Xét:

\(\left(x-z\right)^{2100}=0\hept{\begin{cases}x-z=0\\\frac{5}{3}-z=0\\z=\frac{5}{3}\end{cases}}\)

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

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

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

\(\Leftrightarrow\hept{\begin{cases}3x-5=0\\y^2-1=0\\x-z=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=z=\frac{5}{3}\\y=1\end{cases}}\)

úi! y=1 hoặc y=-1 nha

@@ 

Ta có:

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

Vì \(\left\{{}\begin{matrix}\left(3x-5\right)^{2006}\ge0\\\left(y^2-1\right)^{2008}\ge0\\\left(x-z\right)^{2100}\ge0\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}\left(3x-5\right)^{2006}=0\\\left(y^2-1\right)^{2008}=0\\\left(x-z\right)^{2100}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-5=0\\y^2-1=0\\x-z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}\\y=\pm1\\z=\dfrac{5}{3}\end{matrix}\right.\)

Vậy ...

Chúc bạn học tốt!

=>3x-5=0 và y²-1=0 và x-z=0

=>x=5/3 và y=-1 hoặc y=1 và z=5/3