`a,16(x+4)^{2020}=25(x+4)^{2018}`
`=>16(x+4)^{2020}-25(x+4)^{2018}=0`
`=>(x+4)^{2018}[16(x+4)^2-25]=0`
`=>` $\left[ \begin{array}{l}x+4=0\\16(x+4)^2=25\end{array} \right.$
`=>` $\left[ \begin{array}{l}x=-4\\(x+4)^2=\dfrac{25}{16}\end{array} \right.$
`=>` $\left[ \begin{array}{l}x=-4\\x+4=\dfrac{5}{4}\\x+4=-\dfrac{5}{4}\end{array} \right.$
`=>` $\left[ \begin{array}{l}x=-4\\x=-\dfrac{11}{4}\\x=-\dfrac{21}{4}\end{array} \right.$
Vậy `x=-4` hoặc `x=-11/4` hoặc `x=-21/4`
a) Ta có: \(16\left(x+4\right)^{2020}=25\left(x+4\right)^{2018}\)
\(\Leftrightarrow16\left(x+4\right)^{2020}-25\left(x+4\right)^{2018}=0\)
\(\Leftrightarrow\left(x+4\right)^{2018}\left[16\left(x+4\right)^2-25\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+4\right)^{2018}=0\\16\left(x+4\right)^2-25=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+4=0\\16\left(x+4\right)^2=25\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\\left(x+4\right)^2=\dfrac{25}{16}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x+4=\dfrac{5}{4}\\x+4=-\dfrac{5}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-\dfrac{11}{4}\\x=-\dfrac{21}{4}\end{matrix}\right.\)
Vậy: \(x\in\left\{-4;-\dfrac{11}{4};-\dfrac{21}{4}\right\}\)