A = 2 + 2² + 2³ + 2⁴ + 2⁵..... + 2²³ + 2²⁴

=  (2 + 2² + 2³) + (2³ + 2⁴ + 2⁵) + ..... + (2²² + 2²³ + 2²⁴)

=  (2 + 2² + 2³) + 2² (2 + 2² + 2³) + .... + 2²². (2 + 2² + 2³)

= 14 + 2².14 + .... + 2²².14

= 14.(1 + 2² + ... + 2²²)

= 2.7.(1 + 2² + ... + 2²²) \(⋮\) 7

\(\Rightarrow A⋮7\)(ĐPCM)

a:Sửa đề: \(S=2+2^2+\cdots+2^{2024}\)

Ta có: \(S=2+2^2+\cdots+2^{2024}\)

\(=\left(2+2^2\right)+\left(2^3+2^4\right)+\cdots+\left(2^{2023}+2^{2024}\right)\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+\cdots+2^{2023}\left(1+2\right)\)

\(=3\left(2+2^3+\cdots+2^{2023}\right)\) ⋮3

b: Ta có: \(S=2+2^2+\cdots+2^{2024}\)

\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+\cdots+\left(2^{2021}+2^{2022}+2^{2023}+2^{2024}\right)\)

\(=\left(2+2^2+2^3+2^4\right)+2^4\left(2+2^2+2^3+2^4\right)+\cdots+2^{2020}\left(2+2^2+2^3+2^4\right)\)

\(=30\left(1+2^4+\cdots+2^{2020}\right)=3\cdot10\cdot\left(1+2^4+\cdots+2^{2020}\right)\) ⋮10

=>S có chữ số tận cùng là 0

Lời giải:
$S=(2+2^2)+(2^3+2^4)+....+(2^{23}+2^{24})$

$=2(1+2)+2^3(1+2)+....+2^{23}(1+2)$

$=(1+2)(2+2^3+...+2^{23})$

$=3(2+2^3+...+2^{23})\vdots 3$

b.

$S=2+2^2+2^3+...+2^{23}+2^{24}$

$2S=2^2+2^3+2^4+....+2^{24}+2^{25}$

$\Rightarrow 2S-S=2^{25}-2$

$\Rightarrow S=2^{25}-2$

Ta có:

$2^{10}=1024=10k+4$

$\Rightarrow 2^{25}-2=2^5.2^{20}-2=32(10k+4)^2-2=32(100k^2+80k+16)-2$
$=10(320k^2+8k+51)\vdots 10$

$\Rightarrow S$ tận cùng là $0$

x = 1

Vậy giá trị của x + 224 = 1 + 224 = 225

x+224 có vô số giá trị thỏa mãn đề bài

\(2^x+2^{x+1}+2^{x+2}=224\)

\(\Rightarrow2^x\times\left(1+2+4\right)=224\)

\(\Leftrightarrow2^x\times7=224\)

\(\Leftrightarrow2^x=224\div7\)

\(\Rightarrow2^x=32\)

\(\Leftrightarrow2^x=2^5\)

\(\Rightarrow x=5\)

vậy \(x=5\)

Cam on ban nhieu nha 

giải nhanh giùm mik vs

\(2^x+2^{x+1}+2^{x+2}=224\)

\(7.2^x=224\)

\(\frac{7.2^x}{7}=\frac{224}{7}\)

\(2^x=32\)

\(2^x=2^5\)

\(\Rightarrow x=5\)

Đề có sai ko bn

ko bạn ạ