T làm nhé!
a, \(\left(x-11\right)^{199}=\left(x-11\right)^{200}\)
\(\Leftrightarrow\left(x-11\right)^{200}-\left(x-11\right)^{199}=0\)
\(\Leftrightarrow\left(x-11\right)^{199}\left[\left(x-11\right)-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-11\right)^{199}=0\\x-11-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=12\end{matrix}\right.\)
Vậy \(x=11\) hoặc x = 12
b, \(\left(x-1\right)^2=49^3\)
\(\Leftrightarrow\left(x-1\right)^2=7^6\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=7^3\\x-1=-7^3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=344\\x=-342\end{matrix}\right.\)
Vậy...
Đúng 0
Bình luận (5)
