\(\left(2^3+2\right).n+3^2.n+20=3.5^2\)
\(\left(8+2\right).n+9.n+20=3.25\)
\(10n+9n+20=75\)
\(19n=75-20\)
\(19n=55\)
\(n=55:19=\dfrac{55}{19}\)
Vậy \(n=\dfrac{55}{19}\)
Giải:
\(\left(2^3+2\right).n+3^2.n+20=3.5^2\)
\(\Leftrightarrow n\left(2^3+2+3^2\right)+20=3.5^2\)
\(\Leftrightarrow n\left(8+2+9\right)+20=75\)
\(\Leftrightarrow19n+20=75\)
\(\Leftrightarrow19n=75-20\)
\(\Leftrightarrow19n=55\)
\(\Leftrightarrow n=\dfrac{55}{19}\)
Vậy \(n=\dfrac{55}{19}\).
Chúc bạn học tốt!