\(a,\left(2y+1\right)^3=125\)
\(\Leftrightarrow2y+1=\sqrt[3]{125}=5\)
\(\Leftrightarrow2y=5-1=4\)
\(\Rightarrow y=2\)
\(b,\left(y-5\right)^4=\left(y-5\right)^6\)
\(\Leftrightarrow\left(y-5\right)^4-\left(y-5\right)^6=0\)
\(\Leftrightarrow\left(y-5\right)^4\left[1-\left(y-5\right)^2\right]=0\)
\(\Leftrightarrow\left(y-5\right)^4\left(1-y+5\right)\left(1+y-5\right)=0\)
\(\Leftrightarrow\left(y-5\right)^4\left(6-y\right)\left(y-4\right)=0\)
Vì \(\left(y-5\right)^4>0\forall y\)
\(\Rightarrow\left[{}\begin{matrix}6-y=0\\y-4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=6\\y=4\end{matrix}\right.\)
a) \(\left(2y+1\right)^3=125\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2y+1\right)^3=5^3\\\left(2y+1\right)^3=\left(-5\right)^3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2y+1=5\\2y+1=-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2y=4\\2y=-6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=2\\y=-3\end{matrix}\right.\)
Vậy ...................
