a) Ta có

\(\left\{{}\begin{matrix}3n+1⋮2n+3\\2n+3⋮2n+3\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}6n+2⋮2n+3\\6n+9⋮2n+3\end{matrix}\right.\)

=> 7\(⋮\) 2n + 3

Do n \(\in\) Z nên 2n + 3 \(\in\) Z

=> 2n + 3 \(\in\) Ư₍₇₎ ; 2n + 3 \(⋮̸\) 2

Ta có bảng

Vậy n \(\in\) {-1;2;-2;5} là giá trị cần tìm

a.\(2n^2-3n+1=2n\times\left(n-1\right)-\left(n-1\right)=\left(2n-1\right)\times\left(n-1\right)\Rightarrow2n-1⋮n-1\)

\(\Rightarrow2\left(n-1\right)+1⋮n-1\Rightarrow1⋮n-1\Rightarrow n-1\inƯ\left(1\right)=\left\{1\right\}\Rightarrow n=2\)

b.Tách tương tự nha

\(2n^2-3n+1=\left(2n^2-2n\right)-n+1=2n\left(n-1\right)-n+1\)\(\Rightarrow-n+1⋮n-1\Rightarrow-\left(n-1\right)⋮n-1\)

vậy với mọi x thuộc N đều t/m

b) tương tự nha

a) \(n^2-3n+9\)chia het cho \(n-2\)

\(\Leftrightarrow\)\(n^2-2n-n-2+11\)chia het cho \(n-2\)

\(\Leftrightarrow\)\(\left(n-2\right)\left(n+1\right)+11\)chia het cho \(n-2\)

\(\Leftrightarrow\)11 chia het cho \(n-2\)

\(\Rightarrow\)\(n-2\in U\left(11\right)\)\(\Rightarrow\)\(n-2\in\left\{-11;-1;1;11\right\}\)

                                                   \(\Rightarrow\)\(n\in\left\{-9;1;3;13\right\}\)

b) 2n-1 chia hết cho n-2

\(\Rightarrow2n-2+3\) chia hết cho\(n-2\)

\(\Rightarrow3\)chia hết cho \(n-2\)

\(\Rightarrow n-2\in U\left(3\right)\)\(\Rightarrow n-2\in\left\{-3;-1;1;3\right\}\)\(\Rightarrow n\in\left\{-1;1;3;5\right\}\)

Ta có : \(n+4=n-1+\)\(5\)

Ta thấy : \(\left(n-1\right)⋮\left(n-1\right)\)

Nên \(\left(n+4\right)⋮\left(n-1\right)\Leftrightarrow5⋮\)\(\left(n-1\right)\)

\(\Leftrightarrow\left(n-1\right)\inƯ\left(5\right)=\)\((1;5)\)

a) \(n+4⋮n-1\Rightarrow\left(n-1\right)+5⋮n-1\Rightarrow5⋮n-1\Rightarrow n-1\inƯ\left(5\right)\)

\(\Rightarrow n-1\in\left\{1;5;-1;-5\right\}\Rightarrow n\in\left\{2;6;0;-4\right\}\)

b) \(n^2+2n-3=\left(n^2+n\right)+n-3=n\left(n+1\right)+n-3\)

vì \(n\left(n-1\right)⋮n-1\)\(\Rightarrow n-3⋮n+1\Rightarrow\left(n+1\right)-4⋮n-1\Rightarrow4⋮n-1\Rightarrow n-1\inƯ\left(4\right)\)

\(\Rightarrow n-1\in\left\{1;2;4;-1;-2;-4\right\}\)

\(\Rightarrow n\in\left\{2;3;5;0;-1;-3\right\}\)

\(\left(n^2+2n-3\right)⋮\left(n+1\right)\)

\(n^2+2n-3=\)\(n^2+n+n-3\)

                          \(=n.\left(n+1\right)+n+1-4\)

Mà \(n.\left(n+1\right)⋮\left(n+1\right)\)

           \(\left(n+1\right)⋮\left(n+1\right)\)

Nên \(n^2+2n-3⋮\left(n+1\right)\) \(\Leftrightarrow4⋮\left(n+1\right)\)

                             \(\Leftrightarrow\left(n+1\right)\inƯ\left(4\right)=\)\((1;2;4)\)

Lời giải:

\(M=\frac{1.2.3.4.5.6.7...(2n-1)}{2.4.6...(2n-2).(n+1)(n+2)....2n}=\frac{(2n-1)!}{2.1.2.2.2.3...2(n-1).(n+1).(n+2)...2n}\)

\(=\frac{(2n-1)!}{2^{n-1}.1.2...(n-1).(n+1).(n+2)....2n}=\frac{(2n-1)!}{2^{n-1}.1.2...(n-1).n(n+1)..(2n-1).2}\)

\(=\frac{(2n-1)!}{2^{n-1}.(2n-1)!.2}=\frac{1}{2^{n-1}.2}<\frac{1}{2^{n-1}}\)

Ta có đpcm.

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