a)\(\left(2n+1\right)^3=27\Rightarrow\left(2n+1\right)^3=3^3\)
\(\Rightarrow2n+1=3\)
\(n=\frac{3-1}{2}=\frac{2}{2}=1\)
Vậy \(n=1\)
b)\(\left(n-2\right)^2=n-n\Leftrightarrow\left(n-2\right)^2=0\)
\(\left(n-2\right)^2=0^2\Leftrightarrow n-2=0\)
\(\Leftrightarrow n=0+2=2\)
Vậy \(n=2\)
a) (2n + 1)^3 = 27
=>(2n + 1)^3 = 3^3
=>2n+1=3
=> 2n=2
=>n=1
b) (n-2)^2 = n-n
=>(n-2)^2=0
=>n-2=0
=>n=2
\(a.\left(2n+1\right)^3=27\Rightarrow\left(2n+1\right)^3=3^3\Rightarrow2n+1=3\Rightarrow2n=3-1=2\Rightarrow n=2:2=1\)
\(b.\left(n-2\right)^2=n-n\Rightarrow\left(n-2\right)^2=0\Rightarrow\left(n-2\right)^2=0^2\Rightarrow n-2=0\Rightarrow n=0+2\Rightarrow n=2\)
a) (\(2^n\)+1)\(^3\)=27
3\(^3\)=27
=>n=1
b) n-n=0
=> n=2
