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\(\frac{25}{26}+\frac{1}{26}=\frac{26}{26}=1\)
\(k\)\(mk\)\(nhoa\)
17/45+23/45+28/45+1/27+26/27+22/45
=(17/45+23/45+28/45+22/45)+(1/27+26/27)
=2+1
=3
Điền dấu(<,>.=)
\(\frac{6}{7}...\frac{7}{9}\)\(\frac{31}{32}...\frac{26}{25}\)\(\frac{2014}{4}...1\)
\(\frac{6}{7}\) > \(\frac{7}{9}\)
\(\frac{31}{32}\) < \(\frac{26}{25}\)
\(\frac{2014}{4}\) > 1
Ta có:
\(A=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{3999.4000}}\)
\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{3999}-\frac{1}{4000}}\)
\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\left(1+\frac{1}{3}+...+\frac{1}{3999}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{4000}\right)}\)
\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{3999}+\frac{1}{4000}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{4000}\right)}\)
\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{3999}+\frac{1}{4000}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2000}\right)}\)
\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}=1\)
Ta lại có:
\(B=\frac{\left(17+1\right)\left(\frac{17}{2}+1\right)...\left(\frac{17}{19}+1\right)}{\left(1+\frac{19}{17}\right)\left(1+\frac{19}{16}\right)...\left(1+19\right)}\)
\(=\frac{\frac{18}{1}.\frac{19}{2}.\frac{20}{3}...\frac{36}{19}}{\frac{36}{17}.\frac{35}{16}.\frac{34}{15}...\frac{20}{1}}\)
\(=\frac{1.2.3...36}{1.2.3...36}=1\)
Từ đây ta suy ra được
\(A-B=1-1=0\)
\(\frac{2010}{2011}< \frac{2011}{2012}\)
\(\frac{11}{12}=\frac{22}{24}\)
\(\frac{25}{30}>\frac{25}{49}\)
\(\frac{1}{5}< \frac{3}{8}\)
\(\frac{1995}{1997}< \frac{1995}{1996}\)
\(y.\frac{52}{26}=\frac{17}{13}\)
\(y=\frac{17}{13}:\frac{52}{26}\)
\(y=\frac{17}{26}\)
\(4\frac{1}{2}:y=\frac{2}{7}\)
\(\frac{9}{2}:y=\frac{2}{7}\)
\(y=\frac{9}{2}:\frac{2}{7}\)
\(y=\frac{63}{4}\)
=b chac chan luon