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a: \(\left(n^2+3n-1\right)\left(n+2\right)-n^3+2\)
\(=n^3+2n^2+3n^2+6n-n-2+n^3+2\)
\(=5n^2+5n=5\left(n^2+n\right)⋮5\)
b: \(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-1\right)\)
\(=6n^2+30n+n+5-6n^2+3n-10n+5\)
\(=24n+10⋮2\)
d: \(=\left(n+1\right)\left(n^2+2n\right)\)
\(=n\left(n+1\right)\left(n+2\right)⋮6\)
a: \(A=3^{n+3}+3^{n+1}+2^{n+2}+2^{n+1}\)
\(=3^{n}\cdot27+3^{n}\cdot3+2^{n}\cdot4+2^{n}\cdot2\)
\(=3^{n}\cdot30+2^{n}\cdot6\)
\(=6\left(5\cdot3^{n}+2^{n}\right)\) ⋮6
b:Sửa đề: \(B=3^{n+3}-2^{n+3}+3^{n+1}-2^{n+1}\)
\(=3^{n}\cdot27+3^{n}\cdot3-2^{n}\cdot8-2^{n}\cdot2\)
\(=30\cdot3^{n}-2^{n}\cdot10=10\left(3\cdot3^{n}-2^{n}\right)\) ⋮10
ta có : \(\left(-8\right)^34^{2n}=\left(-2\right)^{3n}.164\Leftrightarrow\left(-2\right)^92^{4n}=\left(-2\right)^{3n}.164\)
\(\Leftrightarrow\dfrac{2^{4n}}{2^{3n}}=\dfrac{164}{2^9}\Leftrightarrow2^n=\dfrac{41}{128}\Leftrightarrow2^{n+7}=41\)
\(\Leftrightarrow n+7=log^{41}_2\Leftrightarrow n=log^{41}_2-7\)
vậy ...
\(\left(-8\right)^3.4^{2n}=\left(-2\right)^{3n}.164\)
đề vậy đúng ko