2n²+5n-1=n(2n-1)+6n-1

             =n(2n-1)+3(2n-1)+2

do 2n²+5n-1 chia hết cho 2n-1 => 2 chia hết cho 2n-1

=> 2n-1 thuộc tập ước của 2 là 1;2

=> n=1 (TM) n=1,5 (loại)

b1

a) \(\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)

\(=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=\dfrac{1}{5}-\dfrac{1}{10}\)

\(=\dfrac{2}{10}-\dfrac{1}{10}\)

\(=\dfrac{1}{10}\)

b) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=\dfrac{1}{1}-\dfrac{1}{100}\)

\(=\dfrac{99}{100}\)

c) \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}\)

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\)

\(=\dfrac{1}{3}-\dfrac{1}{11}\)

\(=\dfrac{8}{33}\)

d) \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

\(=\dfrac{1}{3}-\dfrac{1}{101}\)

\(=\dfrac{98}{303}\)

\(a;\frac{2n+5}{n+3}\)

Gọi \(d\inƯC\left(2n+5;n+3\right)\Rightarrow3n+5⋮d;n+3⋮d\)

\(\Rightarrow2n+5⋮d\)và \(2\left(n+3\right)⋮d\)

\(\Rightarrow\left[\left(2n+6\right)-\left(2n+5\right)\right]⋮d\)

\(\Rightarrow1⋮d\Rightarrow d=1\)

Vậy \(\frac{2n+5}{n+3}\)là phân số tối giản

\(B=\frac{2n+5}{n+3}=\frac{2\left(n+3\right)+5-6}{n+3}=\frac{2\left(n+3\right)-1}{n+3}=2-\frac{1}{n+3}\)

Với \(B\in Z\)để n là số nguyên 

\(\Rightarrow1⋮n+3\Rightarrow n+3\inƯ\left(1\right)=\left\{\pm1\right\}\)

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

Vậy.....................

a, \(\frac{2n+5}{n+3}\)Đặt \(2n+5;n+3=d\left(d\inℕ^∗\right)\)

\(2n+5⋮d\) ; \(n+3⋮d\Rightarrow2n+6\)

Suy ra : \(2n+5-2n-6⋮d\Rightarrow-1⋮d\Rightarrow d=1\)

Vậy tta có đpcm 

b, \(B=\frac{2n+5}{n+3}=\frac{2\left(n+3\right)-1}{n+3}=\frac{-1}{n+3}=\frac{1}{-n-3}\)

hay \(-n-3\inƯ\left\{1\right\}=\left\{\pm1\right\}\)

{ 300 - [25.2^2 + 2^2.5]:2}

={300 -[25.4 + 4.5]:2}

={ 300 - 120 : 2}

= { 300 - 60}

= 240

2^2.(2x +1) = 5

<=> 4.(2x +1) = 5

<=> 2x + 1 = 1,25

<=> 2x = 2,25

<=> x = 1,125

Mình sẽ chấm cho 10 người đấu tiên nhé😻😻😻😻a

\(B=\frac{1^2}{2^2-1}.\frac{3^2}{4^2-1}.\frac{5^2}{6^2-1}...\frac{\left(2n+1\right)^2}{\left(2n+2\right)^2-1}\)

\(=\frac{1^2}{\left(2-1\right)\left(2+1\right)}.\frac{3^2}{\left(4-1\right)\left(4+1\right)}...\frac{\left(2n+1\right)^2}{\left(2n+2-1\right)\left(2n+2+1\right)}\)

\(=\frac{1}{1.3}.\frac{3^2}{3.5}...\frac{\left(2n+1\right)^2}{\left(2n+1\right)\left(2n+3\right)}\)

\(=\frac{1}{2n+3}\)

a) \(1+2+3+4+...+n\)

\(=\left(n+1\right)\left[\left(n-1\right):1+1\right]:2\)

\(=\left(n+1\right)\left(n-1+1\right):2\)

\(=n\left(n+1\right):2\)

\(=\dfrac{n\left(n+1\right)}{2}\)

b) \(2+4+6+..+2n\)

\(=\left(2n+2\right)\left[\left(2n-2\right):2+1\right]:2\)

\(=2\left(n+1\right)\left[2\left(n-1\right):2+1\right]:2\)

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

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

c) \(1+3+5+...+\left(2n+1\right)\)

\(=\left[\left(2n+1\right)+1\right]\left\{\left[\left(2n-1\right)-1\right]:2+1\right\}:2\)

\(=\left(2n+1+1\right)\left[\left(2n-1-1\right):2+1\right]:2\)

\(=\left(2n+2\right)\left[\left(2n-2\right):2+1\right]:2\)

\(=2\left(n+1\right)\left[2\left(n-1\right):2+1\right]:2\)

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

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

d) \(1+4+7+10+...+2005\)

\(=\left(2005+1\right)\left[\left(2005-1\right):3+1\right]:2\)

\(=2006\cdot\left(2004:3+1\right):2\)

\(=2006\cdot\left(668+1\right):2\)

\(=1003\cdot669\)

\(=671007\)

e) \(2+5+8+...+2006\)

\(=\left(2006+2\right)\left[\left(2006-2\right):3+1\right]:2\)

\(=2008\cdot\left(2004:3+1\right):2\)

\(=1004\cdot\left(668+1\right)\)

\(=1004\cdot669\)

\(=671676\)

g) \(1+5+9+...+2001\)

\(=\left(2001+1\right)\left[\left(2001-1\right):4+1\right]:2\)

\(=2002\cdot\left(2000:4+1\right):2\)

\(=1001\cdot\left(500+1\right)\)

\(=1001\cdot501\)

\(=501501\)

a) \(2+4+6+...+2n=n\left(n+1\right)\)       (1)

\(n=1\) ta có : \(2=1\cdot\left(1+1\right)\)  ( đúng)

Giả sử (1) đúng đến n, ta sẽ chứng minh (1) đúng với n+1

Có \(2+4+6+...+2n+2\left(n+1\right)\)

\(=n\left(n+1\right)+2\left(n+1\right)=\left(n+1\right)\left(n+2\right)\)

=> (1) đúng với n+1

Theo nguyên lý quy nạp ta có đpcm

b) sai đề nha, mình search google thì được như này =))

 \(1^3+3^3+5^3+...+\left(2n-1\right)^2=n^2\left(2n^2-1\right)\)     (2)

\(n=1\) ta có : \(1^3=1^2\cdot\left(2-1\right)\)   (đúng) 

giả sử (2) đúng đến n, tức là \(1^3+3^3+...+\left(2n-1\right)^3=n^2\left(2n^2-1\right)\)

Ta c/m (2) đúng với n+1

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

\(=2n^4+8n^3+11n^2+6n+1\)

\(=\left(n^2+2n+1\right)\left(2n^2+4n+1\right)\)

\(=\left(n+1\right)^2\left[2\left(n+1\right)^2-1\right]\)   => (2) đúng với n+1

Theo nguyên lý quy nạp ta có đpcm