Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
P=(-1)n.(-1)2n+1.(-1)n+1
=(-1)n.(-1)n.(-1)n+1.(-1)n+1
=(-1)2n.(-1)2n+2
=1.1( vì 2n;2n+2 đều là số chẵn)
=1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\left(-1\right)^n.\left(-1\right)^{2n+1}.\left(-1\right)^{n+1}\)
=\(\left(-1\right)^{n+2n+1+n+1}=\left(-1\right)^{4n+2}\)
ta thấy 4n+2 là mũ chẵn nên P=1
![](https://rs.olm.vn/images/avt/0.png?1311)
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: \(=\left(n+1\right)\left(n^2+2n\right)\)
\(=n\left(n+1\right)\left(n+2\right)⋮6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(P=\left(x-4\right)^{\left(x-5\right)^{\left(x-6\right)^{\cdot\left(x+6\right)^{\left(x+5\right)}}}}\)
Thay x = 7 ta được:
\(P=\left(7-4\right)^{\left(7-5\right)^{\left(7-6\right)^{\left(7+6\right)^{\left(7+5\right)}}}}\)
\(P=3^{2^{1^{13^{12}}}}=3^2.1^{13^{12}}=9.1=9\)
b, Vì \(x-1=x-1\) nên lũy thừa của nó phải giống nhau
mà \(x+2\ne x+4\)
\(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\) có nghiệm \(\Leftrightarrow\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-2\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\frac{1.3.5...\left(2n-1\right)}{\left(n+1\right).\left(n+2\right).\left(n+3\right)...2n}=\frac{\left(1.3.5...2n-1\right).\left(2.4.6...2n\right)}{\left(2.4.6...2n\right)\left(n+1\right).\left(n+2\right).\left(n+3\right)...2n}\)
\(=\frac{1.2.3.4.5.6...\left(2n-1\right).2n}{1.2.3...n\left(n+1\right).\left(n+2\right).\left(n+3\right)...2n.2^n}\)
\(=\frac{1}{2^n}\)
\(\Rightarrow P=\left(-1\right)^{n+2n+1+n+1}=\left(-1\right)^{4n+2}=\left(-1\right)^{2.\left(2n+1\right)}=1^{2n+1}=1\)