Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
G = \(\frac{2^2}{1.3}\).\(\frac{3^2}{2.4}\).\(\frac{4^2}{3.5}\).....\(\frac{50^2}{49.51}\)
=> G = \(\frac{2.2}{1.3}\).\(\frac{3.3}{2.4}\).\(\frac{4.4}{3.5}\).....\(\frac{50.50}{49.51}\)
=> G = \(\frac{2.2.3.3.4.4.....50.50}{1.2.3.3.4.4.....50.51}\)
=> G = \(\frac{2.50}{1.51}\)
=> G = \(\frac{100}{51}\)
Bài 1 :
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{9}{19}\)
\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{9}{19}\)
\(\Leftrightarrow1-\frac{1}{2x+3}=\frac{9}{19}\)
\(\Leftrightarrow\frac{1}{2x+3}=1-\frac{9}{19}\)
\(\Leftrightarrow\frac{1}{2x+3}=\frac{10}{19}\)
\(\Leftrightarrow10.\left(2x+3\right)=19\Leftrightarrow2x+3=\frac{19}{10}\)
\(\Leftrightarrow2x=\frac{19}{10}-3\Leftrightarrow2x=-\frac{11}{10}\)
\(\Leftrightarrow x=-\frac{11}{20}=-0,55\)
Bài 2 :
\(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2016.2018}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{2016}-\frac{1}{2018}\)
\(=\frac{1}{2}-\frac{1}{2018}=\frac{504}{1009}\)
1/2.(1/3+1/6+1/10+...+1/x(x+1))=1/2.2016/2018
1/6+1/12+1/20+...+1/x(x+1)=504/1009
1/2.3+1/3.4+1/4.5+...+1/x(x+1)=504/1009
1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1=504/1009
1/2-1/x+1=504/1009
x-1/2(x+1)=504/1009
-> 1009(x-1)=504.2(x+1)
1009x-1009=1008x+1008
1009x-1008x=1008+1009
->x=2017
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}=\frac{2016}{2018}\)
\(A=\frac{1}{2\left(2+1\right):2}+\frac{1}{3\left(3+1\right):2}+...+\frac{1}{x\left(x+1\right):2}\)
\(A=\frac{1}{2\left(2+1\right)}\cdot2+\frac{1}{3\left(3+1\right)}\cdot2+...+\frac{1}{x\left(x+1\right)}.2=\frac{2016}{2018}\)
\(A=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2016}{2018}\)
\(A=2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2016}{2018}\)
\(A=1-\frac{1}{x+1}=\frac{2016}{2018}\)
\(\Rightarrow\frac{1}{x+1}=1-\frac{2016}{2018}=\frac{1}{1009}\)
\(\Rightarrow x+1=1009\Rightarrow x=1008\)
\(A=4+\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{19}\right)\cdot\left(1-\frac{1}{20}\right)\)
\(A=4+\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{18}{19}\cdot\frac{19}{20}\right)\)
\(A=4+\frac{1\cdot2\cdot3\cdot...\cdot18\cdot19}{2\cdot3\cdot4\cdot...\cdot19\cdot20}\)
\(A=4+\frac{1}{20}\)
\(A=\frac{81}{20}\)
= ?
Mk ko biet