Câu 2: 

#include <bits/stdc++.h>

using namespace std;

double p1,p2;

int i,n;

int main()

{

cin>>n;

p1=1;

p2=1;

for (i=1; i<=n; i++)

{

if (i%2==0) p2=p2*(i*1.0);

else p1=p1*(i*1.0);

}

cout<<fixed<<setprecision(2)<<p1<<endl;

cout<<fixed<<setprecision(2)<<p2;

return 0;

}

1:

\(\lim\limits_{n\rightarrow\infty}\dfrac{3n^5+3n^3-1}{n^3-2n}=\lim\limits_{n\rightarrow\infty}\dfrac{n^5\left(3+\dfrac{3}{n^2}-\dfrac{1}{n^5}\right)}{n^3\left(1-\dfrac{2}{n^2}\right)}\)

\(=\lim\limits_{n\rightarrow\infty}n^2\cdot3=+\infty\)

2: \(\lim\limits_{n\rightarrow\infty}\dfrac{3n^7+3n^5-n}{3n^2-2n}=\lim\limits_{n\rightarrow\infty}\dfrac{3n^6+3n^4-1}{3n-2}\)

\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^6\left(3+\dfrac{3}{n^2}-\dfrac{1}{n^6}\right)}{n\left(3-\dfrac{2}{n}\right)}=\lim\limits_{n\rightarrow\infty}n^5=+\infty\)

Bạn sử dụng cách tính tổng của một dãy số cách đều. 
Giải: 
1+3+5+...+(2n-1)=225 
<=>{[(2n-1)+1].[(2n-1)-1]:2 + 1}/2 = 225 
<=> (2n.2n):4 = 225 
<=> n^2=225 
suy ra n = 15 và n = -15 
do n thuộc N* nên n = 15 thỏa mãn

tick mình nha

\(\frac{\left(\frac{-5}{7}\right)^{n+1}}{\left(\frac{-5}{7}\right)^n}\)

\(=\frac{-5}{7}\)

#)Giải :

\(\frac{\left(\frac{-5}{7}\right)^{n+1}}{\left(\frac{-5}{7}\right)^n}=\frac{\left(\frac{-5}{7}\right)^n\times\left(\frac{-5}{7}\right)}{\left(\frac{-5}{7}\right)^n}=\frac{-5}{7}\)

\(\frac{\left(\frac{-5}{7}\right)^{n+1}}{\left(\frac{-5}{7}\right)^n}=\left(\frac{-5}{7}\right)^{n+1-n}=\left(\frac{-5}{7}\right)^1=\frac{-5}{7}\)

a, \(2^{-1}.2^n+4.2^n=9.2^5\)

\(\Rightarrow2^n.\frac{9}{2}=288\)

\(\Rightarrow2^n=64\)

\(\Rightarrow n=6\)

\(KL....\)

b, đề hơi sai pn ạ

c, \(7^6+7^5-7^4=7^4\left(7^2+7-1\right)=7^4.55\)chia hết cho 55

d, \(A=1+5+5^2+5^3+...+5^{49}+5^{50}\)

\(\Rightarrow5A=5+5^2+5^3+5^4+...+5^{50}+5^{51}\)

\(\Rightarrow5A-A=5^{51}-1\)

\(\Rightarrow A=\frac{5^{51}-1}{4}\)

a, 2−1.2n+4.2n=9.25

⇒2n.92 =288

⇒2n=64

⇒n=6

KL....

b, đề hơi sai pn ạ

c, 76+75−74=74(72+7−1)=74.55chia hết cho 55

d, A=1+5+52+53+...+549+550

⇒5A=5+52+53+54+...+550+551

⇒5A−A=551−1

⇒A=551−14