\(3.2^{x+1}-5.2^{13}=7.2^{13}\\ =>3.2^{x+1}=7.2^{13}+5.2^{13}\\ =>3.2^{x+1}=2^{13}\left(7+5\right)\\ =>3.2^{x+1}=2^{13}.12\\ =>2^{x+1}=2^{13}.4\\ =>2^{x+1}=2^{13}.2^2\\ =>2^{x+1}=2^{15}\\ =>x+1=15\\ =>x=14\)

Ta có :

3.2^x+1-5.2¹³=7.2¹³

=> 3.2^x+1=7.2¹³+5.2¹³

=> 3.2^x+1=(7+5).2¹³

=> 3.2^x+1=12.2¹³

=> 2^x+1=4.2¹³

=> 2^x+1=2².2¹³

=> 2^x+1=2¹⁵

=> x+1=15

=> x=14

a, gọi x là số tự nhiên thỏa mãn đề bài ta có 

x ⋮2, 3, 4, 5 ⇔ x ϵ BC(2,3,4,5)

2 = 2; 3 =3; 4 = 2.2; 5 = 5 ⇔ BCNN(2,3,4,5) = 2.3.2.5 = 60

⇔ x ϵ B(60) ⇔ x = 60.k , k ϵ N*

vậy số tự nhiên chia hết cho 2,3,4,5 có dạng tổng quát là 

x = 60k với k ϵ N*

help me]

a)Ta có :

A=2019.2011=(2010-1).(2010+1)

=2010.2010+2010.1-1.2010-1.1

= 2010²-1<2010²=B

b)Ta có :

A=2015.2017=(2016-1).(2016+1)

= 2016.2016+2016.1-2016.1-1.1

= 2016²-1<2016²=B

a số 2 lớn hơn

\(\left(x+1\right)⋮x+1\)

\(\left(x+1\right)+2⋮x+1\\ \)

\(⇌\left(x+1\right)\inƯ\left(2\right)\)

\(Ư\left(2\right)=\left\{1;-1;2;-2\right\}\)

=> \(x=\left\{0;-2;1;-3\right\}\)

\(\left(x-2021\right)\left(x-2022\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2021=0\\x-2022=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2021\\x=2022\end{matrix}\right.\)

Vậy ...

X=0

\(x-1\in\text{Ư}\left(28\right)\)

\(\Rightarrow x-1\in\left\{1;-1;2;-2;4;-4;7;-7;14;-14\right\}\)

\(x^{2022}=x^{2021}\\ =>x^{2022}-x^{2021}=0\\ =>x^{2021}\left(x-1\right)=0\\ =>\left[{}\begin{matrix}x^{2021}=0\\x-1=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

x²⁰²²=x²⁰²¹

=> x²⁰²²-x²⁰²¹=0

=> x²⁰²¹.(x-1)=0

\(\Rightarrow\left[{}\begin{matrix}x^{2021}=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

`24-(x-16)=14`

`x-16=24-14=10`

`x=10+16`

`x=26`

\(24-\left(x-16\right)=14\\ =>x-16=24-14\\ =>x-16=10\\ =>x=10+16\\ =>x=26\)