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a)\(\frac{1313}{1515}=\frac{13\times101}{15\times101}=\frac{13}{15}\)
\(\frac{1326}{1428}=\frac{1326:102}{1428:102}=\frac{13}{14}\)
DO \(\frac{13}{14}>\frac{13.}{15}\) nên \(\frac{1326}{1428}>\frac{1313}{1515}\)
\(b\))\(\frac{222}{555}\) và \(\frac{333}{444}\)
\(\frac{222}{555}=\frac{2\times111}{5\times111}=\frac{2}{5}\)
\(\frac{333}{444}=\frac{3\times111}{4\times111}=\frac{3}{4}\)
DO \(\frac{2}{5}< \frac{3}{4}\) nên \(\frac{222}{555}< \frac{333}{444}\)
Ta có: \(\frac{1212}{1515}=\frac{12}{15}\)
=>\(\frac{12}{15}=\frac{1212}{1515}\)
\(\frac{1717}{3232}=\frac{17}{32};\frac{1515}{1616}=\frac{15}{16}\)
\(\frac{1515}{1717}\) \(=\) \(\frac{1515:101}{1717:101}\) \(=\) \(\frac{15}{17}\)
\(\frac{625}{600}\) \(=\) \(\frac{625:25}{600:25}\) \(=\) \(\frac{25}{24}\)
Chúc bạn học tốt !
a. Ta có : \(\frac{1313}{1515}=\frac{13.101}{15.101}=\frac{13}{15}\)
\(\frac{1326}{1428}=\frac{13.102}{14.102}=\frac{13}{14}\)
Vì \(\frac{13}{15}< \frac{13}{14}\)(do 15> 14)
=>\(\frac{1313}{1515}< \frac{1326}{1428}\)
b.Ta có : \(1-\frac{119}{120}=\frac{1}{120}\)
\(1-\frac{118}{119}=\frac{1}{119}\)
Vì \(\frac{1}{120}< \frac{1}{119}\)(do 120>119)
=> \(1-\frac{1}{120}>1-\frac{1}{119}\)
=>\(\frac{119}{120}>\frac{118}{119}\)
Mình giải theo kiểu lớp 6 nhá !
=\(33.\left(\frac{34}{15}+\frac{34}{35}+\frac{34}{63}+\frac{34}{99}\right)\)
=\(33.\left[34.\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\right]\)
=\(33.\left[34.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)\right]\)
=\(33.\left[34.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right).\frac{1}{2}\right]\)
=\(33.\left[34\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right).\frac{1}{2}\right]\)
=\(33.\left[34.\left(\frac{1}{3}-\frac{1}{11}\right).\frac{1}{2}\right]\)
=\(33.\left(34.\frac{8}{33}.\frac{1}{2}\right)\)
=\(33.\frac{136}{33}\)
=\(\frac{33.136}{33}\)(*)
=\(136\)
(Bạn có thể bỏ bước có dấu *)
\(\frac{1515}{1515}=1\)
\(\frac{2000}{1999}>1\)
\(\Rightarrow\frac{1515}{1515}< \frac{2000}{1999}\)
vì 1515/1515=1 và 2000/1999>1