b, \(\left(5x+1\right)^2=\frac{36}{49}\)

\(\Rightarrow\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\)

\(\Rightarrow5x+1=\frac{6}{7}\)

\(\Rightarrow5x=\frac{-1}{7}\)

\(\Rightarrow x=\frac{-1}{35}\)

#include <bits/stdc++.h>

using namespace std;

int main()

{

long long a,c,tam;

cin>>a>>c;

cout<<"a truoc khi doi la:"<<a<<endl;

cout<<"c truoc khi doi la:"<<c<<endl;

tam=a;

a=c;

c=tam;

cout<<"a sau khi doi la:"<<a<<endl;

cout<<"c sau khi doi la:"<<c<<endl;

return 0;

}

a) 1 + 3 + 5 + ... + 13

= (13 + 1).[(13 - 1) : 2 + 1] : 2

= 14 . 7 : 2

= 49

= 7²

b) 3² + 4² + 12²

= 9 + 16 + 144

= 169

= 13²

\(\left(\dfrac{1}{4}\right)^{2n}=\left(\dfrac{1}{8}\right)^2\)

\(\Rightarrow\left(\dfrac{1}{2}\right)^{2.2n}=\left(\dfrac{1}{2}\right)^{3.2}\)

\(\Rightarrow\left(\dfrac{1}{2}\right)^{4n}=\left(\dfrac{1}{2}\right)^6\)

\(\Rightarrow4n=6\)

\(\Rightarrow n=\dfrac{6}{4}=\dfrac{3}{2}\)

n = 3/2

1/2 mũ n = 1/2 mũ 6

Vậy x = 6

\(\left(\dfrac{1}{2}\right)^n=\left(\dfrac{1}{8}\right)^2\)

\(=>\left(\dfrac{1}{2}\right)^n=\left[\left(\dfrac{1}{2}\right)^3\right]^2\)

\(=>\left(\dfrac{1}{2}\right)^n=\left(\dfrac{1}{2}\right)^6\)

\(\Rightarrow n=6\)

\(\left(\dfrac{1}{2}\right)^n=\left(\dfrac{1}{8}\right)^2 \Rightarrow\left(\dfrac{1}{2}\right)^n=\dfrac{1}{64} \Rightarrow\left(\dfrac{1}{2}\right)^n=\left(\dfrac{1}{2}\right)^6\Rightarrow n=6\)

1. 2^x=16\(\Rightarrow\)X=4

2. 2^2x-1=2⁷

\(\Rightarrow\)2⁷=2^2.4-1

Vậy x =4

x=4 nha chị

Bài 3: 

1: =>2x=-5/3-1/2=-10/6-3/6=-13/6

hay x=-13/12

2: =>3/5x=1/7+3/5=5/35+21/35=26/35

hay x=26/3

3: =>-3x=5/6+3/4=10/12+9/12=19/12

hay x=-19/36

4: =>1/2x=3/7-5/4=12/28-35/28=-23/28

hay x=-23/14

5: =>1/4x=-3/5-7/5=-2

hay x=-8

6: =>3x=1/42+1/7=1/42+6/42=1/7

hay x=1/21