giải giúp mình với
1. \(\left(2n+1\right)^2=\dfrac{1}{25}\)
<=> \(\left(2n+1\right)^2=\left(\dfrac{1}{5}\right)^2\)
<=> \(\left|2n+1\right|=\left|\dfrac{1}{5}\right|\)
<=> \(\left[{}\begin{matrix}2n+1=\dfrac{1}{5}\\2n+1=-\dfrac{1}{5}\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}n=-\dfrac{2}{5}\\n=-\dfrac{3}{5}\end{matrix}\right.\)
2. \(\left(3n-2\right)^3=\dfrac{27}{48}\)
<=> \(\left(3n-2\right)^3=\left(\dfrac{3}{\sqrt[3]{48}}\right)^3\)
<=> \(\left|3n-2\right|=\left|\dfrac{3}{\sqrt[3]{48}}\right|\)
<=> \(\left[{}\begin{matrix}3n-2=\dfrac{3}{\sqrt[3]{48}}\\3n-2=-\dfrac{3}{\sqrt[3]{48}}\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}n=0,94...\\n=0,39...\end{matrix}\right.\)
a: \(\left(2n+1\right)^2=\dfrac{1}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}2n+1=\dfrac{1}{5}\\2n+1=-\dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2n=\dfrac{-4}{5}\\2n=\dfrac{-6}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}n=\dfrac{-2}{5}\\n=\dfrac{-3}{5}\end{matrix}\right.\)
)
