\(12,8\times\dfrac{1}{2}+12,8\times0,25+12,8\times\dfrac{1}{4}\)

\(=12,8\times0,5+12,8\times0,25+12,8\times0,25\)

\(=12,8\times\left(0,5+0,25+0,25\right)\)

\(=12,8\times1\)

\(=12,8\)

\(12,8\times\dfrac{1}{2}+12,8\times0,25+12,8\times\dfrac{1}{4}\\ =12,8\times\dfrac{1}{2}+12,8\times\dfrac{25}{100}+12,8\times\dfrac{1}{4}\\ =12,8\times\dfrac{1}{2}+12,8\times\dfrac{1}{4}+12,8\times\dfrac{1}{4}\\ =12,8\times\left(\dfrac{2}{4}+\dfrac{1}{4}+\dfrac{1}{4}\right)\\ =12,8\times\dfrac{4}{4}=12,8\times1=12,8\)

Sửa đề : \(\dfrac{2025\times2024-1}{2023\times2025+2024}\)

\(=\)\(\dfrac{2025\times\left(2023+1\right)-1}{2023\times2025+2024}\)

\(=\dfrac{2025\times2023+2025-1}{2023\times2025+2024}\)

\(=\dfrac{2025\times2023+\left(2025-1\right)}{2023\times2025+2024}\)

\(=\dfrac{2025\times2023+2024}{2023\times2025+2024}\)

\(=1\)

@\(\text{格雷斯}\)

Đặt \(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...+\dfrac{1}{1024}\)

\(2\times A=2\times\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...+\dfrac{1}{1024}\right)\)

\(=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{512}\)

\(2\times A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{512}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...+\dfrac{1}{1024}\right)\)

\(A=1-\dfrac{1}{1024}=\dfrac{1023}{1024}\)

\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times...\times\left(1-\dfrac{1}{2023}\right)\times\left(1-\dfrac{1}{2024}\right)\)

\(=\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times...\times\dfrac{2022}{2023}\times\dfrac{2023}{2024}\)

\(=\dfrac{1\times2\times3\times...\times2022\times2023}{2\times3\times4\times...\times2023\times2024}\)

\(=\dfrac{1}{2024}\)

\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times...\times\left(1-\dfrac{1}{2023}\right)\times\left(1-\dfrac{1}{2024}\right)\)

\(=\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times...\times\dfrac{2022}{2023}\times\dfrac{2023}{2024}\)

\(=\dfrac{1\times2\times3\times...\times2022\times2023}{2\times3\times4\times...\times2023\times2024}\)

\(=\dfrac{1}{2024}\)

\(Z=\dfrac{3}{3\times5}+\dfrac{3}{5\times7}+\dfrac{3}{7\times9}+...+\dfrac{3}{49\times51}\)

\(=\dfrac{3}{2}\times\left(\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+...+\dfrac{2}{49\times51}\right)\)

\(=\dfrac{3}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)

\(=\dfrac{3}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{51}\right)\)

\(=\dfrac{3}{2}\times\dfrac{16}{51}=\dfrac{8}{17}\)

\(Z=\dfrac{3}{3x5}+\dfrac{3}{5x7}+\dfrac{3}{7x9}+...+\dfrac{3}{49x51}\\ =\dfrac{3}{2}x\left(\dfrac{2}{3x5}+\dfrac{2}{5x7}+\dfrac{2}{7x9}+...+\dfrac{2}{49x51}\right)\\ =\dfrac{3}{2}x\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\\ =\dfrac{3}{2}x\left(\dfrac{1}{3}-\dfrac{1}{51}\right)\\ =\dfrac{3}{2}x\dfrac{16}{51}=\dfrac{8}{17}\)

\(S=\dfrac{2}{1\times2}+\dfrac{2}{2\times3}+\dfrac{2}{3\times4}+...+\dfrac{2}{99\times100}\)

\(=2\times\left(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{99\times100}\right)\)

\(=2\times\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(=2\times\left(1-\dfrac{1}{100}\right)\)

\(=2\times\dfrac{99}{100}=\dfrac{99}{50}\)

CT: \(\dfrac{a}{n\left(n+a\right)}=\dfrac{1}{n}-\dfrac{1}{n+a};\left(n\ne0;n\ne-a\right)\)

\(S=\dfrac{2}{1\times2}+\dfrac{2}{2\times3}+\dfrac{2}{3\times4}+...+\dfrac{2}{99\times100}\\ \dfrac{S}{2}=\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{99\times100}\\ \dfrac{S}{2}=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ \dfrac{S}{2}=1-\dfrac{1}{100}=\dfrac{99}{100}\\ S=\dfrac{99}{100}\times2=\dfrac{99}{50}\)

\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{9\times10}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)

CT: \(\dfrac{a}{n\left(n+a\right)}=\dfrac{1}{n}-\dfrac{1}{n+a}\) (\(n\ne0;n\ne-a\))

\(\dfrac{1}{1x2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+...+\dfrac{1}{9x10}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-...-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\\ =1-\dfrac{1}{10}\\ =\dfrac{9}{10}\)

Diện tích hình tròn là:

   \(3,14\times25\times25:4=490,625\left(cm^2\right)\)

          Đáp số: \(490,625cm^2\)

Bán kính hình tròn đó là :

\(25\div2=12,5\left(cm\right)\)

S hình tròn đó là :

\(12,5\times12,5\times3,14=490,625\left(cm^2\right)\)

Bài toán được mô tả như hình sau:

Sửa đề: chuyển cm thành m

Độ dài đáy bé được mở rộng thêm là:

\(600:\dfrac{1}{2}:30=40\left(m\right)\)

Độ dài đáy bé của thửa ruộng là:

\(160-40=120\left(m\right)\)

Diện tích thửa ruộng là:

\(\dfrac{\left(160+120\right)\times30}{2}=4200\left(m^2\right)=0,42\left(ha\right)\)

Năm 2023 số tiền hoa anh Nam thu được trên thửa ruộng đó là:

\(0,42\times650000000=273000000\) (đồng)

Dòng chữ TOQUOCVIETNAM có 13 chữ cái và lặp đi lặp lại

Do 2018 : 13 = 155 (dư 3) nên chữ cái thứ 2018 là chữ cái thứ 3 trong dãy TOQUOCVIETNAM nên là chữ Q

Chọn C