Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có : \(\frac{3}{5}+\frac{6}{11}+\frac{7}{13}+\frac{2}{5}+\frac{5}{11}+\frac{19}{13}+\frac{1}{2}+\frac{2}{3}-\frac{1}{6}\)
\(=\left(\frac{3}{5}+\frac{2}{5}\right)+\left(\frac{6}{11}+\frac{5}{11}\right)+\left(\frac{7}{13}+\frac{19}{13}\right)+\left(\frac{1}{2}+\frac{2}{3}-\frac{1}{6}\right)\)
\(=1+1+2+1=5\)
= (3/5 + 2/5) + (6/11 + 5/11) + (7/13 + 19/13) : 1/2 + 2/3 - 1/6
= (1 + 1 + 2) : (1/2 + 4/6 - 1/6)
= 4 : (1/2 + 1/2)
= 4 : 1
= 4
Vậy kết quả bằng 4
\(\frac{6}{7}+\frac{2}{5}-\frac{1}{2}+2\frac{38}{91}-\frac{4}{5}=\frac{27300}{31850}+\frac{12740}{31850}-\frac{15925}{31850}+\) \(2\frac{13300}{31850}\)\(-\frac{25480}{31850}\)
\(=2\frac{27300+12740-15925+13300-25480}{31850}\)\(=2\frac{341}{910}\)
HỌC TỐT!!!
Gọi biểu thức trên là \(A\)
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{256}\)
\(2A=\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{256}\right)\times2\)
\(2A=\frac{1}{2}\times2+\frac{1}{4}\times2+\frac{1}{8}\times2+\frac{1}{16}\times2+...+\frac{1}{512}\times2\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}\right)\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{128}-\frac{1}{256}\)
\(A=1-\frac{1}{256}\)
\(A=\frac{255}{256}\)
\(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{110}\)
= \(\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+...+\dfrac{1}{10\times11}\)
= \(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{10}-\dfrac{1}{11}\)
= \(\dfrac{1}{3}-\dfrac{1}{11}=\dfrac{8}{33}\)
viết rõ lại cái đề giùm tui tui mới làm được
1/2+1/6+1/12+1/20+...+1/380
=1/1.2+ 1/2.3+ 1/3.4+ 1/4.5+...+ 1/19.20
= 1/1-1/2+ 1/2-1/3+ 1/3-1/4+ 1/4-1/5+...+1/19-1/20
=1-1/29
=29/29-1/29
= 28/29
28/29