Thuế VAT mà bác Minh phải trả khi mua chiếc điện thoại là:

\(10\%\cdot7990000=799000\left(đ\right)\) 

Số tiền mà bác Minh phải trả khi mua chiếc điện thoại là:

\(7990000+799000=8789000\left(đ\right)\) 

$20+(4^2-4)-2$

$=(20-2)+(16-4)$

$=18+12=30$

30

2B:

a: \(A=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

\(=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}\cdot\dfrac{\dfrac{3}{4}\left(1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}\right)}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

\(=\dfrac{1}{2}\cdot\dfrac{3}{4}+\dfrac{5}{8}=\dfrac{3}{8}+\dfrac{5}{8}=\dfrac{8}{8}=1\)

b: \(B=\dfrac{0,125-\dfrac{1}{5}+\dfrac{1}{7}}{0,375-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0,2}{\dfrac{3}{4}+0,5-\dfrac{3}{10}}\)

\(=\dfrac{\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}}{3\left(\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}\right)}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{6}}{\dfrac{3}{2}\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{6}\right)}=\dfrac{1}{3}+\dfrac{2}{3}=1\)

a) \(A=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

\(=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}\cdot\dfrac{\dfrac{3}{4}\left(1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}\right)}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

\(=\dfrac{1}{2}\cdot\dfrac{3}{4}+\dfrac{5}{8}\)

\(=\dfrac{3}{8}+\dfrac{5}{8}=1\)

b) \(B=\dfrac{0,125-\dfrac{1}{5}+\dfrac{1}{7}}{0,375-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0,2}{\dfrac{3}{4}+0,5-\dfrac{3}{10}}\)

\(=\dfrac{\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}}{\dfrac{3}{8}-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{4}+\dfrac{1}{2}-\dfrac{3}{10}}\)

\(=\dfrac{\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}}{3\left(\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}\right)}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{2}\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}\right)}\)

\(=\dfrac{1}{3}+\dfrac{1}{\dfrac{3}{2}}=\dfrac{1}{3}+\dfrac{2}{3}=1\)

\(B=\left(1-\dfrac{1}{2^2}\right)\cdot\left(1-\dfrac{1}{3^2}\right)\cdot\left(1-\dfrac{1}{4^2}\right)\cdot...\cdot\left(1-\dfrac{1}{2024^2}\right)\)

\(=\dfrac{2^2-1}{2^2}\cdot\dfrac{3^2-1}{3^2}\cdot\dfrac{4^2-1}{4^2}\cdot...\cdot\dfrac{2024^2-1}{2024^2}\)

Ta có CT: \(a^2-1=\left(a+1\right)\left(b+1\right)\)

\(B=\dfrac{\left(2+1\right)\left(2-1\right)}{2^2}\cdot\dfrac{\left(3+1\right)\left(3-1\right)}{3^2}\cdot\dfrac{\left(4+1\right)\left(4-1\right)}{4^2}...\cdot\dfrac{\left(2024+1\right)\left(2024-1\right)}{2024^2}\) 

\(=\dfrac{1\cdot3}{2^2}\cdot\dfrac{4\cdot2}{3^2}\cdot\dfrac{5\cdot3}{4^2}\cdot...\cdot\dfrac{2025\cdot2023}{2024^2}\)

\(=\dfrac{1\cdot2\cdot3^2\cdot...\cdot2023^2\cdot2024\cdot2025}{2^2\cdot3^2\cdot...\cdot2024^2}\)

\(=\dfrac{2025}{2\cdot2024}=\dfrac{2025}{4048}>\dfrac{2024}{4048}=\dfrac{1}{2}\)

Vậy: ...  

Ta có : 

\(B=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right).....\left(1-\dfrac{1}{2024^2}\right)\)

\(=\dfrac{2^2-1}{2^2}.\dfrac{3^2-1}{3^2}.\dfrac{4^2-1}{4^2}.....\dfrac{2024^2-1}{2024^2}\)

\(=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}.....\dfrac{2023.2025}{2024^2}\)

\(=\dfrac{1.2.3.....2023}{2.3.4.....2024}.\dfrac{3.4.5.....2025}{2.3.4.....2024}\)

\(=\dfrac{1}{2024}.\dfrac{2025}{2}=\dfrac{2025}{4048}>\dfrac{1}{2}\)

Vậy \(B>\dfrac{1}{2}\)

x+(x+1)+(x+2)+...+(x+10)=505

11x+(1+2+...+10)=505

11x+[(10+1).10:2]=505

11x+55=505

11x=450

x=\(\dfrac{450}{11}\)

Vậy \(x=\dfrac{450}{11}\)

x + (x + 1) + (x + 2) + ... + (x + 10) = 505 

(x + x + ... + x) + (1 + 2 + 3 + ... + 10) = 505 

11x + 55 = 505

11x = 505 - 55

11x = 450

x = 450 : 11

x = `450/11` 

\(\dfrac{2x-34}{2}=14\Rightarrow2x-34=28\Leftrightarrow2x=62\Leftrightarrow x=31\)

x = 31

\(\dfrac{x-4}{2020}+\dfrac{x-3}{2021}+\dfrac{x-2}{2022}+\dfrac{x-1}{2023}+\dfrac{x-2024}{5}=4\) (sửa đề)

\(\Rightarrow\left(\dfrac{x-4}{2020}-1\right)+\left(\dfrac{x-3}{2021}-1\right)+\left(\dfrac{x-2}{2022}-1\right)+\left(\dfrac{x-1}{2023}-1\right)+\dfrac{x-2024}{5}=0\)

\(\Rightarrow\dfrac{x-2024}{2020}+\dfrac{x-2024}{2021}+\dfrac{x-2024}{2022}+\dfrac{x-2024}{2023}+\dfrac{x-2024}{5}=0\)

\(\Rightarrow\left(x-2024\right)\left(\dfrac{1}{2020}+\dfrac{1}{2021}+\dfrac{1}{2022}+\dfrac{1}{2023}+\dfrac{1}{5}\right)=0\)

\(\Rightarrow x-2024=0\) (vì \(\dfrac{1}{2020}+\dfrac{1}{2021}+\dfrac{1}{2022}+\dfrac{1}{2023}+\dfrac{1}{5}\ne0\))

\(\Rightarrow x=2024\)

\(\dfrac{x-4}{2020}-1+\dfrac{x-3}{2021}-1+\dfrac{x-2}{2022}-1+\dfrac{x-1}{2023}-1+\dfrac{x-2024}{5}+2=0\)

\(\Leftrightarrow\dfrac{x-2024}{2020}+\dfrac{x-2024}{2021}+\dfrac{x-2024}{2022}+\dfrac{x-2024}{2023}+\dfrac{x-2024}{5}+2=0\)

\(\Leftrightarrow\left(x-2024\right)\left(\dfrac{1}{2020}+\dfrac{1}{2021}+\dfrac{1}{2022}+\dfrac{1}{2023}+\dfrac{1}{5}\right)+2=0\)

\(\Leftrightarrow x=-\dfrac{2}{\dfrac{1}{2020}+\dfrac{1}{2021}+\dfrac{1}{2022}+\dfrac{1}{2023}+\dfrac{1}{5}}+2024\)

SỬA ĐỀ: b) Chứng tỏ S>n-2... & Điều kiện: \(n\inℕ^∗\) và \(n>2\) (theo quy luật)

a) \(S=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+...+\dfrac{n^2-1}{n^2}\) 

\(S=\left(1-\dfrac{1}{4}\right)+\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{1}{16}\right)+...+\left(1-\dfrac{1}{n^2}\right)\)

\(S=1-\dfrac{1}{2^2}+1-\dfrac{1}{3^2}+1-\dfrac{1}{4^2}+...+1-\dfrac{1}{n^2}\)

\(S=n-1-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}\right)\)

Nhận xét: 

\(n-1-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}\right)< n-1\)

\(\Rightarrow S< n-1\) (*)

b) Nhận xét:

\(\left\{{}\begin{matrix}\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}\\\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}\\...\\\dfrac{1}{n^2}< \dfrac{1}{\left(n-1\right)\cdot n}\end{matrix}\right.\)

\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{\left(n-1\right)\cdot n}=1-\dfrac{1}{n}< 1\)

\(\Rightarrow-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}\right)>1\)

\(\Rightarrow n-1-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}\right)>n-1-1=n-2\)

\(\Rightarrow S>n-2\) (**) 

Từ (*) và (**) suy ra:

\(n-2< S< n-1\)

Mà \(n-1\) và \(n-2\) là 2 số tự nhiên liên tiếp nên:

S không thể là một số tự nhiên 

Vậy S không thể là một số tự nhiên 

Số lượng các chữ cái khác nhau trong nhóm từ "nghìn năm thăng long" là:

Chữ N: 5 chữ

Chữ G: 3 chữ 

Chữ H: 2 chữ

Chữ I: 1 chữ

Chữ Ă: 2 chữ

Để dãy từ "nghìnnămthăng longnghìnnămthănglong..." khi viết thành chữ số được một số lớn nhất thì các chữ cái khác nhau sẽ tương ứng với các chữ số khác nhau là:
Chữ N = 9

Chữ G = 8

Chữ H = 7 

Chữ I = 6

Chữ Ă = 5

Do đó tổng các chữ số trong nhóm "nghìn năm thăng long" là:

\(\left(9\cdot5\right)+\left(8\cdot3\right)+\left(7\cdot2\right)+\left(6\cdot1\right)+\left(5\cdot2\right)=99\)

Suy ra tổng các chữ số trong dãy "nghìnnămthăng longnghìnnămthănglong..." là một số chia hết cho 99

Mà \(1102010⋮̸99\) nên tổng các chữ số trong dãy "nghìnnămthăng longnghìnnămthănglong..." không thể là 1102010

Vậy...

Mình chỉ làm theo ý mình hiểu nên không biết đúng không

\(\dfrac{\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}}{\dfrac{3}{8}-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{15}{30}+\dfrac{10}{30}-\dfrac{6}{30}}{\dfrac{15}{20}+\dfrac{10}{20}-\dfrac{6}{20}}\)

\(=\dfrac{\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}}{3\left(\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}\right)}+\dfrac{\left(15+10-6\right)\cdot\dfrac{1}{30}}{\left(15+10-6\right)\cdot\dfrac{1}{20}}\)

\(=\dfrac{1}{3}+\dfrac{1}{30}:\dfrac{1}{20}=\dfrac{1}{3}+\dfrac{2}{3}=1\)