Tính giá trị biểu thức:
\(\left(-1\right)^{2n}\left(-1\right)^n\left(-1\right)^{n+1}\left(n\in Z\right)\)
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Ta có : \(\left(n-1\right)\left(3-2n\right)-n\left(n+5\right)\)
\(=n\left(3-2n\right)-\left(3-2n\right)-n^2-5n\)
\(=3n-2n^2-3+2n-n^2-5n\)
\(=-3n^2-3\)
\(=-3\left(n^2+1\right)⋮3\)
Vậy \(\left(n-1\right)\left(3-2n\right)-n\left(n+5\right)⋮3\)
Ta có \(\left(n-1\right)\left(3-2n\right)-n\left(n+5\right)=3n-2n^2-3+2n-n^2-5n=-3n-3\)
mà -3n chia hết cho 3,-3 chia hết cho 3
=> biểu thức (n-1)(3-2n)-n(n+5) chia hết cho 3(đpcm)
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
\(\left(-2\right).\left(-1\frac{1}{2}\right)\left(-1\frac{1}{3}\right).....\left(-1\frac{1}{2013}\right)\)
\(=\left(-2\right).\left(\frac{-3}{2}\right)\left(-\frac{4}{3}\right)......\left(\frac{-2014}{2013}\right)\)
\(=\frac{\left(-2\right).\left(-3\right).\left(-4\right)....\left(-2014\right)}{2.3.....2013}\)
\(=\frac{2.3.4....2014\left(\text{Vì có 2014 thừa số âm }\right)}{2.3....2013}\)
\(=\frac{\left(2.3.4....2013\right).2014}{2.3....2013}\)
\(=2014\)
1. 2n-3 ⋮ n+1
⇒2n+2-5 ⋮ n+1
⇒2(n+1)-5 ⋮ n+1
Do n∈Z
⇒n+1 ∈ Ư(-5)={-1,1,-5,5}
⇒\(\left[{}\begin{matrix}n-1=-1\\n-1=1\\n-1=-5\\n-1=5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}n=0\\n=2\\n=-4\\n=6\end{matrix}\right.\)
Vậy x∈{0,2,-4,6}
2. Ta có:
x-y-z=0 ⇒\(\left\{{}\begin{matrix}x=y+z\\y=x-z\\z=x-y\end{matrix}\right.\)
Thay vào biểu thức ta được:
\(B=\left(1-\frac{x-y}{x}\right)\left(1-\frac{y+z}{y}\right)\left(1+\frac{x-z}{z}\right)\)
⇒\(B=\frac{x-x+y}{x}.\frac{y-y-z}{y}.\frac{z+x-z}{z}\)
⇒\(B=\frac{y.\left(-z\right).x}{x.y.z}=\frac{\left(-1\right)xyz}{xyz}=-1\)
Vậy biểu thức B có giá trị là -1
\(A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdot\cdot\cdot\left(1-\frac{1}{n^2}\right)\)
\(\Rightarrow A=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\cdot\cdot\cdot\left(1-\frac{1}{n^2}\right)\)
\(\Rightarrow A=\frac{3}{4}\cdot\frac{8}{9}\cdot\cdot\cdot\frac{n^2-1}{n^2}\)
\(\Rightarrow A=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot\cdot\cdot\frac{\left(n-1\right)\left(n+1\right)}{n\cdot n}\)
\(\Rightarrow A=\frac{\left(1\cdot3\right)\cdot\left(2\cdot4\right)\cdot\cdot\cdot\left[\left(n-1\right)\left(n+1\right)\right]}{\left(2\cdot2\right)\cdot\left(3\cdot3\right)\cdot\cdot\cdot\left(n\cdot n\right)}\)
\(\Rightarrow A=\frac{\left[1\cdot2\cdot\cdot\cdot\cdot\cdot\left(n-1\right)\right]\cdot\left[3\cdot4\cdot\cdot\cdot\cdot\cdot\left(n+1\right)\right]}{\left(2\cdot3\cdot\cdot\cdot\cdot\cdot n\right)\cdot\left(2\cdot3\cdot\cdot\cdot\cdot\cdot n\right)}\)
\(\Rightarrow A=\frac{1\cdot\left(n+1\right)}{n\cdot2}\)
\(\Rightarrow A=\frac{n+1}{2n}\)
A=(1-1/2^2)(1-1/3^2).....(1-1/n^2)
A=1(1/2^2-1/3^2-...-1/n^2)
......
xin lỗi bạn nha mình phải tắt máy rồi bạn cố gắng suy nghĩ tiếp nha
\(U_n=\dfrac{\left(n^2-1\right)}{n\left(n+2\right)}U_{n-1}\Rightarrow n\left(n+2\right).U_n=\left(n-1\right)\left(n+1\right).U_{n-1}\)
Đặt \(n\left(n+2\right).U_n=V_n\Rightarrow V_{n-1}=\left(n-1\right)\left(n+2-1\right).U_{n-1}=\left(n-1\right).\left(n+1\right)U_{n-1}\)
\(\Rightarrow V_n=V_{n-1}\)
\(\Rightarrow V_n=V_{n-1}=V_{n-2}=...=V_1\)
Có \(V_1=1.\left(1+2\right).U_1=1\)
\(\Rightarrow V_n=1\)
\(\Rightarrow U_n=\dfrac{V_n}{n\left(n+2\right)}=\dfrac{1}{n\left(n+2\right)}\)
\(\Rightarrow A=\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+...+\dfrac{1}{2015.2017}\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\right)\)
\(=\dfrac{1}{2}\left(1+\dfrac{1}{2}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)\)
\(=...\)
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: \(=\left(n+1\right)\left(n^2+2n\right)\)
\(=n\left(n+1\right)\left(n+2\right)⋮6\)
\(\left(-1\right)^{2n}\left(-1\right)^n\left(-1\right)^{n+1}=\left(-1\right)^{2n+n+n+1}=\left(-1\right)^{3n+1}\)