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\(\frac{6}{7}+\frac{2}{5}-\frac{1}{2}+2\frac{38}{91}-\frac{4}{5}=\frac{27300}{31850}+\frac{12740}{31850}-\frac{15925}{31850}+\) \(2\frac{13300}{31850}\)\(-\frac{25480}{31850}\)
\(=2\frac{27300+12740-15925+13300-25480}{31850}\)\(=2\frac{341}{910}\)
HỌC TỐT!!!
viết rõ lại cái đề giùm tui tui mới làm được
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
\(\dfrac{4}{5}\)+\(\dfrac{7}{10}\) -\(\dfrac{1}{2}\)
Giữ nguyên phân số \(\dfrac{7}{10}\) ( MSC: 10)
\(\dfrac{4}{5}\)= \(\dfrac{4x2}{5x2}\) =\(\dfrac{8}{10}\)
\(\dfrac{1}{2}\) = \(\dfrac{1x5}{2x5}\) = \(\dfrac{5}{10}\)
\(\dfrac{4}{5}\)+\(\dfrac{7}{10}\) -\(\dfrac{1}{2}\) = \(\dfrac{8}{10}\) +\(\dfrac{7}{10}\) -\(\dfrac{5}{10}\) =1
Gọi biểu thức trên là \(A\)
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{256}\)
\(2A=\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{256}\right)\times2\)
\(2A=\frac{1}{2}\times2+\frac{1}{4}\times2+\frac{1}{8}\times2+\frac{1}{16}\times2+...+\frac{1}{512}\times2\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}\right)\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{128}-\frac{1}{256}\)
\(A=1-\frac{1}{256}\)
\(A=\frac{255}{256}\)
Ta có: \(\frac{\frac{6}{7}+\frac{2}{5}-\frac{1}{2}}{\frac{4}{5}+\frac{5}{6}-\frac{1}{3}}+2\frac{38}{91}\)
\(=\frac{\frac{53}{70}}{\frac{13}{10}}+\frac{220}{91}\)
\(=\frac{53}{91}+\frac{220}{91}=3\)
TL
=13/10 NHA BN
HT
K CHO MIK ĐI