`Answer:`

Đặt \(d=ƯCLN\left(2n+3;4n+7\right)\)

\(\Rightarrow\hept{\begin{cases}2n+3⋮d\\4n+7⋮d\end{cases}\Rightarrow\hept{\begin{cases}2\left(2n+3\right)⋮d\\4n+7⋮d\end{cases}}\Rightarrow\hept{\begin{cases}4n+6⋮d\\4n+7⋮d\end{cases}}}\)

\(\Rightarrow\left(4n+7\right)-\left(4n+6\right)⋮d\)

\(\Rightarrow1⋮d\)

\(\Rightarrow d=\pm1\)

Vậy `\frac{2n+3}{4n+7}` tối giản ` ∀n`

-29 phần 15 nhé

cần mn giúp cảm ơn :D

a) \(n\in Z;n\ne-1\)

b) Để A là số nguyên thì \(6⋮n+1\) (\(n\in Z;n\ne-1\))

\(\Rightarrow\)\(n+1\inƯ\left(6\right)\)

\(\Rightarrow\)\(n+1\in\left\{1;2;3;6;-1;-2;-3;-6\right\}\)

Vì \(n\in Z;n\ne-1\).Ta có bảng:

Thử lại:Đúng

Vậy \(n\in\left\{0;1;2;-2;-3;-4;-7;5\right\}\)

Chúc bạn hok giỏi nha!

B=1/1.2+1/3.4+1/5.6+...+1/99.100 

=1‐1/2+1/3‐1/4+1/5‐1/6+...+1/99‐1/100 

=﴾1+1/3+1/5+...+1/99﴿‐﴾1/2+1/4+1/6+...+1/100﴿ 

=﴾1+1/2+1/3+1/4+1/5+1/6+...+1/99+1/100﴿‐2﴾1/2+1/4+1/6+...+1/100﴿ 

=﴾1+1/2+1/3+1/4+...+1/100﴿‐﴾1+1/2+1/3+..+1/50﴿ 

=1/51+1/52+1/53+..+1/100 ﴾1﴿ 

A=1/51+1/52+1/53+..+1/100 ﴾2﴿ 

Từ ﴾1﴿,﴾2﴿=> A/B=1 

HT

`Answer:`

`\frac{x-3}{4}=\frac{4}{x-3}`

`<=>(x-3)(x-3)=4.4`

`<=>(x-3)^2=16`

`<=>x-3=-4` hoặc `x-3=4`

`<=>x=-1` hoặc `x=7`