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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: \(=\left(n+1\right)\left(n^2+2n\right)\)
\(=n\left(n+1\right)\left(n+2\right)⋮6\)
\(\frac{1}{n^3}< \frac{1}{\left(n-2\right)n\left(n+1\right)}\Leftrightarrow\frac{\left(n-2\right)n\left(n+1\right)}{n^3}< 1\Leftrightarrow\left(n-2\right)\left(n+1\right)< n^2\)
\(\Leftrightarrow n^2-n-2< n^2\Leftrightarrow-n-2< 0\)Đúng \(\forall n\inℕ\)
--->ĐPCM
\(\sqrt{1+2+3...+\left(n-1\right)+n+\left(n-1\right)+...+3+2+1}\)
\(=\sqrt{2\left[1+2+3+..+\left(n-1\right)+n\right]}=\sqrt{2\frac{n\left(n-1\right)}{2}+n}\)
\(=\sqrt{n\left(n-1\right)+n}=\sqrt{n^2-n+n}=\sqrt{n^2}=n\left(đpcm\right)\)
\(\sqrt{1+2+3+...+\left(n-1\right)+n+\left(n-1\right)+...+3+2+1}\\ =\sqrt{2\left[1+2+3+...+\left(n-1\right)+n\right]-n}\\ =\sqrt{2.\left(n+1\right).n:2-n}\\ =\sqrt{n\left(n+1\right)-n}\\ =\sqrt{n^2+n-n}\\ =\sqrt{n^2}\\ =n\)