A=-6/5x?viết sai rùi bạn : 4/5

à, bạn j ấy ơi, đề mình đc giao là như thế đấy ạ không nhầm đâu nha!

Ta có : \(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\)

\(\Leftrightarrow2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{18.19.20}\)

\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{18.19}-\frac{1}{19.20}\)

\(=\frac{1}{2}-\frac{1}{19.20}=\frac{189}{380}\)

\(\Rightarrow B=\frac{189}{760}\)

\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{18.19.20}\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}\left(\frac{1}{18.19}-\frac{1}{19.20}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{19.20}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{380}\right)\)

\(=\frac{1}{2}.\frac{189}{380}=\frac{189}{760}\)

\(B=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{18.19.20}\right).\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right).\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{380}\right)\)

\(=\frac{179}{760}\)

P=1/1.2.3.4 +1/2.3.4.5 +1/3.4.5.6 +...+1/97.98.99.100 

3P=3/1.2.3.4 +3/2.3.4.5 +3/3.4.5.6 +...+3/97.98.99.100

3P=1/1.2.3-1/2.3.4+1/2.3.4-1/3.4.5+................+1/97.98.99-1/98.99.100

3P = 1/1.2.3 - 1/98.99.100

3P =( 98.99.100-1.2.3)/1.2.3.98.99.100

P=( 98.99.100-1.2.3)/1.2.3.98.99.100.3

P=(98.33.50-1)/98.99.100.3

P= 161699/2910600

=398759

mk cũng cần lắm, ở trong toán vui mỗi tuần trên online math

Bài 1  :

\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2018}}{\frac{2017}{1}+\frac{2016}{2}+\frac{2015}{3}+...+\frac{1}{2017}}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}{\left(\frac{2017}{1}+1\right)+\left(\frac{2016}{2}+1\right)+\left(\frac{2015}{3}+1\right)+...+\left(\frac{1}{2017}+1\right)+1}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}{\frac{2018}{1}+\frac{2018}{2}+\frac{2018}{3}+....+\frac{2018}{2017}+\frac{2018}{2018}}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}{2018.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}+\frac{1}{2018}\right)}\)

\(=\frac{1}{2018}\)

B=\(\frac{\frac{1}{51}+\frac{1}{53}+...+\frac{1}{149}}{\frac{1}{101.99}+\frac{1}{103.97}+...+\frac{1}{149.51}}\)

\(\)TA CÓ E=\(\frac{1}{101.99}+\frac{1}{103.97}+...+\frac{1}{149.51}\)

\(200E=\frac{200}{101.99}+\frac{200}{103.97}+..+\frac{200}{149.51}\)

\(200E=\frac{101+99}{101.99}+\frac{103+97}{103.97}+...+\frac{149+51}{149.51}\)

\(200E=\frac{1}{99}+\frac{1}{101}+\frac{1}{97}+\frac{1}{103}+...+\frac{1}{51}+\frac{1}{149}\)

\(200E=\frac{1}{51}+\frac{1}{53}+...+\frac{1}{147}+\frac{1}{149}\)

\(E=\left(\frac{1}{51}+\frac{1}{53}+...+\frac{1}{147}+\frac{1}{149}\right):200\)\(=\left(\frac{1}{51}+\frac{1}{53}+...+\frac{1}{147}+\frac{1}{149}\right).\frac{1}{200}\)

\(\Rightarrow B=\frac{1}{51}+\frac{1}{53}+...+\frac{1}{149}\)/\(\left(\frac{1}{51}+\frac{1}{53}+..+\frac{1}{149}\right).\frac{1}{200}\)

\(\Rightarrow B=\frac{1}{\frac{1}{200}}=200\)

VẬY B=200

Còn câu C thì bạn làm tương tự câu B thôi bạn nhé

                       \(M=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}.\) Nhân với 2 cả hai vế:

được:          \(2M=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)  Suy ra :     \(M=2M-M=1-\frac{1}{2^{100}}\)

CHÚC BẠN HỌC GIỎI

=>A:1/2=1/1x3+1/3x5+1/5x7+...+1/99x101

=>2a=1/2(2/1x3+2/3x5+...+2/99x101)

từ đây tự làm

\(A=\frac{1}{2.3}+\frac{1}{6.5}+\frac{1}{10.7}+...+\frac{1}{198.101}\)

\(\Rightarrow2A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

\(\Rightarrow2A=\frac{1}{2}\left(1-\frac{1}{101}\right)\)

\(\Rightarrow4A=\frac{100}{101}\)

\(\Leftrightarrow A=\frac{100}{101}.\frac{1}{4}=\frac{4.25}{101.4}=25< 26\)