Olm chào em, em chưa hiểu chỗ nào em nhỉ?

\(\dfrac{2}{3}-\left(-\dfrac{1}{4}\right)+\dfrac{3}{5}-\dfrac{7}{45}-\left(-\dfrac{5}{9}\right)+\dfrac{1}{12}+\dfrac{1}{35}\\ =\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{7}{45}+\dfrac{5}{9}+\dfrac{1}{12}+\dfrac{1}{35}\\ =\left(\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{1}{12}\right)+\left(\dfrac{3}{5}+\dfrac{5}{9}-\dfrac{7}{45}\right)+\dfrac{1}{35}\\ =\left(\dfrac{8}{12}+\dfrac{3}{12}+\dfrac{1}{12}\right)+\left(\dfrac{27}{45}+\dfrac{25}{45}-\dfrac{7}{45}\right)+\dfrac{1}{35}\\ =\dfrac{12}{12}+\dfrac{45}{45}+\dfrac{1}{35}\\ =1+1+\dfrac{1}{35}\\ =2+\dfrac{1}{35}\\ =\dfrac{70}{35}+\dfrac{1}{35}=\dfrac{71}{35}\)

\(\dfrac{2}{3}-\left(-\dfrac{1}{4}\right)+\dfrac{3}{5}-\dfrac{7}{45}-\left(-\dfrac{5}{9}\right)+\dfrac{1}{12}+\dfrac{1}{35}\)

\(=\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{7}{45}+\dfrac{5}{9}+\dfrac{1}{12}+\dfrac{1}{35}\)

\(=\left(\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{1}{12}\right)+\left(\dfrac{3}{5}+\dfrac{1}{35}\right)+\left(-\dfrac{7}{45}+\dfrac{5}{9}\right)\)

\(=\left(\dfrac{8}{12}+\dfrac{3}{12}+\dfrac{1}{12}\right)+\left(\dfrac{21}{35}+\dfrac{1}{35}\right)+\left(-\dfrac{7}{45}+\dfrac{25}{45}\right)\)

\(=1+\dfrac{22}{35}+\dfrac{18}{45}\)

\(=\dfrac{315}{315}+\dfrac{198}{315}+\dfrac{126}{315}\)

\(=\dfrac{71}{35}\)

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:

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

Trong sữa có \(3 , 6 \%\) bơ. Tính lượng sữa trong một chai, biết rằng lượng bơ trong chai sữa này là \(14 , 4\) g.

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

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

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 \(B>\dfrac{1}{2}\)

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

A = {\(x\) = 2k + 1/ k\(\in\) N; 6≤ k ≤ 14}

B = {\(x\)  = 2k/ k \(\in\) N; 11 ≤ k ≤ 21}

D = {\(x\) = k²/ k \(\in\) N; 2 ≤ k ≤ 7}

A={x\(\in\)N|13<=x<=29; \(x=2k+1;k\in N\)}

B={x\(\in\)N|22<=x<=42: \(x⋮\)2}

C={x\(\in\)N|7<=x<=29; \(x=4k+3\left(k\in N\right)\)}

D={x\(\in\)N|\(4< =x< =49;x=k^2;k\in N\)}

\(x\in\) N; \(x\) + 3 = 10;

           \(x\) + 3 = 10

           \(x\)        = 10 - 3

           \(x\) = 7

C = {7}

C = {7}

Diện tích mảnh đất là:

\(50\times8=400\left(m^2\right)\)

Diện tích đất để xây nhà là:

\(400\times25\%=100\left(m^2\right)\)

Vậy...

diện tích mãnh đất hình chữ nhật là 

\(S=50.8=400m^2\)

100% vuông tương ứng 400 mét vuông nên 25% tương ứng 

\(\dfrac{25.400}{100}=100m^2\)

vậy diện tích đất đai để xây nhà là 100 mét vuông

3x-7=x-7

=> 3x-x=7-7

=> 2x=0

=> x=0:2

=> x=0

x=0