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c; 17\(\dfrac{2}{31}\) - (\(\dfrac{15}{17}\) + 6\(\dfrac{2}{31}\))
= 17 + \(\dfrac{2}{31}\) - \(\dfrac{15}{17}\) - 6 - \(\dfrac{2}{31}\)
= (17 - 6) - \(\dfrac{15}{17}\) + (\(\dfrac{2}{31}\) - \(\dfrac{2}{31}\))
= 11 - \(\dfrac{15}{17}\)+ 0
= \(\dfrac{172}{17}\)
b; 130\(\dfrac{25}{28}\) + 120\(\dfrac{17}{35}\)
= 130 + \(\dfrac{25}{28}\) + 120 + \(\dfrac{17}{35}\)
= (130 + 120) + (\(\dfrac{25}{28}\) + \(\dfrac{17}{35}\))
= 250 + (\(\dfrac{125}{140}\) + \(\dfrac{68}{140}\))
= 250 + \(\dfrac{193}{140}\)
= 250\(\dfrac{193}{140}\)
Các hỗn só và phân số = nhau là:5/3=1 12/18; 5 9/31=1 29/65; 55/14=5 13/44; 94/65=164/31;2 7/13=66/26
Ta có: x-(\(\frac{31}{5}+\frac{31}{3.5}+\frac{31}{5.7}+\frac{31}{7.9}+\frac{31}{9.11}\)\(+\frac{31}{11.13}\))=\(\frac{9}{13}\)
x-\(\frac{31}{5}\)-\(\frac{31}{2}\)x(\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\))=\(\frac{9}{13}\)
x-\(\frac{31}{5}-\frac{31}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{11}-\frac{1}{13}\right)\)=\(\frac{9}{13}\)
x-\(\frac{31}{5}\)\(-\frac{31}{2}\left(\frac{1}{3}-\frac{1}{13}\right)=\frac{9}{13}\)
x-\(\frac{31}{5}-\frac{31}{2}.\frac{10}{39}\)\(=\frac{9}{13}\)
x-\(\frac{31}{5}-\frac{155}{39}=\frac{9}{13}\)
x-\(\frac{434}{195}\)=\(\frac{9}{13}\)
x =\(\frac{9}{13}+\frac{434}{195}=\frac{569}{195}\)
nhé
A=1+1/2+1+1/6+1+1/12+...+1+1/90=
=9+1/2+1/6+1/12+...+1/90
1/2+1/6+1/12+...+1/90=
1/1x2+1/2x3+2/3x4+...+1/9x10=
\(=\dfrac{2-1}{1x2}+\dfrac{3-2}{2x3}+\dfrac{4-3}{3x4}+...+\dfrac{10-9}{9x10}=\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}=\)
\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)
\(\Rightarrow A=9+\dfrac{9}{10}=9\dfrac{9}{10}\)
\(\left(31\frac{6}{13}+5\frac{9}{31}\right)-30\frac{6}{13}=31\frac{6}{13}-30\frac{6}{13}+5\frac{9}{31}\)
\(=1+5\frac{9}{31}=6\frac{9}{31}\)