Đề bài sai phải ko???

Đặt A = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\)

- Vì : 

 \(\frac{1}{2^2}< \frac{1}{1.2}\)

\(\frac{1}{3^2}< \frac{1}{2.3}\)

...................

\(\frac{1}{n^2}< \frac{1}{n\left(n-1\right)}\)

Cộng vế với vế , ta suy ra 

A < \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}\)= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.....+\frac{1}{n-1}-\frac{1}{n}\)

= \(1-\frac{1}{n}< 1\)

=> A<1 ( đpcm )

Ta có:\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\)>\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\)=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\)=\(\frac{1}{1}-\frac{1}{n}\)<1 => \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< 1\)

Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< 1\)

Ta có : \(4n+5⋮5\)

\(\Leftrightarrow4n⋮5\)

\(\Leftrightarrow n⋮5\)

\(\Rightarrow n\inℕ\left(ĐK:n\in B_{\left(5\right)}\right)\)

\(b,3n+4⋮n-1\)

Ta có : \(\frac{3n+4}{n-1}=\frac{3n-3+7}{n-1}=\frac{3(n-1)+7}{n-1}=3+\frac{7}{n-1}\)

Do đó : \(7⋮n-1\)=> \(n-1\inƯ(7)\)

=> \(n-1\in\left\{1;7\right\}\)

=> \(n\in\left\{2;8\right\}\)

a, \(4n+5⋮5\)

\(\Rightarrow4n+5\in\text{Ư}\left(5\right)=\left\{1;5\right\}\)

\(\Rightarrow\orbr{\begin{cases}4n+5=1\\4n+5=5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}4n=-4\\4n=0\end{cases}}\Rightarrow\orbr{\begin{cases}n=-1\left(\text{loại}\text{ vì n thuộc N }\right)\\n=0\end{cases}}\)

Vậy n = 0 

b, \(3n+4⋮n-1\)

\(\Rightarrow3\left(n-1\right)+7⋮n-1\)

MÀ \(3\left(n-1\right)⋮n-1\)

Nên \(7⋮n-1\)hay \(n-1\in\text{Ư}\left(7\right)=\left\{1;7\right\}\)

\(\Rightarrow\orbr{\begin{cases}n-1=1\\n-1=7\end{cases}\Rightarrow\orbr{\begin{cases}n=2\\n=8\end{cases}}}\)

Vậy n = 2 hoặc n  = 8 

a) 25^{n + 2} : 5^{n + 1}

 = (5²)^{(n + 2)} : 5^{n + 1}

 = 5^{2.(n + 2)}  : 5^{n + 1}

 = 5^{2n + 4} : 5^{n + 1}

= 5^{2n + 4 - (n + 1)}

= 5^{n + 3}

b) 8^{n + 5} : 4^{n + 1}

= (2³)^{(n + 5)} : (2²)^{(n + 1)}

= 2^{3(n + 5)} : 2^{2(n + 1)}

= 2^{3n + 15} : 2^{2n + 2}

= 2^{3n + 15 - (2n + 2)}

= 2^{n + 13}

đề câu c cũng như câu a thôi bạn