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Mn tách từng phần ra cho dễ tl nhá !Mơn trc
a, \(\frac{3}{8}+\frac{11}{13}-\frac{9}{13}\)
=\(\frac{3}{8}+\frac{2}{13}\)
=\(\frac{55}{104}.\)
b, \(\frac{2}{7}.\left(\frac{5}{9}+\frac{4}{9}\right)+\frac{2}{7}\)
=\(\frac{2}{7}.\frac{9}{9}+\frac{2}{7}\)
=\(\frac{2}{7}+\frac{2}{7}\)
=\(\frac{4}{7}\)
c, \(\frac{3}{11}.\left(\frac{3}{5}-\frac{5}{3}\right)-\frac{3}{10}.\left(\frac{1}{3}-\frac{2}{5}\right)\)
=\(\frac{3}{11}.-\frac{16}{15}-\frac{3}{10}.-\frac{1}{15}\)
=\(-\frac{16}{55}--\frac{1}{50}\)
=\(-\frac{149}{550}.\)
d, \(\frac{-3}{4}.\frac{11}{23}+\frac{3}{23}.\frac{31}{17}-\frac{3}{17}.\frac{19}{23}\)
=\(-\frac{33}{92}+\frac{93}{391}-\frac{57}{391}\)
=\(-\frac{417}{1564}\)
e, \(\frac{3}{17}.\frac{11}{23}+\frac{3}{23}.\frac{31}{17}-\frac{3}{17}.\frac{19}{23}\)
=\(\frac{33}{391}+\frac{93}{391}--\frac{254}{391}\)
=\(\frac{380}{391}.\)
g, \(\frac{3}{7}.\frac{-5}{12}+\frac{11}{17}:\frac{5}{-12}\)
=\(-\frac{5}{28}+-\frac{132}{85}\)
= \(-1.731512605.\)
k cho mình nha làm mỏi tay quá ,.....................kết bạn với mình nha.......................
a) \(\frac{31}{23}-\left(\frac{7}{32}+\frac{8}{23}\right)=\frac{31}{23}-\frac{7}{32}-\frac{8}{23}=1-\frac{7}{32}=\frac{25}{32}\)
b) \(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
\(=\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}\)
\(=\frac{1}{3}-\left(\frac{79}{67}-\frac{12}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)\)
\(=\frac{1}{3}-1+1=\frac{1}{3}\)
d) \(\frac{1}{7}.\frac{1}{3}+\frac{1}{7}.\frac{-1}{3}+\frac{17}{19}=\frac{1}{7}.\left(\frac{1}{3}-\frac{1}{3}\right)+\frac{17}{19}=\frac{17}{19}\)
e) \(\frac{3}{5}.\frac{7}{9}+\frac{7}{5}.\frac{2}{9}=\frac{7}{5}.\left(\frac{3}{9}+\frac{2}{9}\right)=\frac{7}{5}.\frac{5}{9}=\frac{7}{9}\)
a)\(11\frac{1}{4}-\left(2\frac{5}{7}+5\frac{1}{4}\right)\)
\(=\frac{45}{4}-\left(\frac{19}{7}+\frac{21}{4}\right)\)
\(=\frac{45}{4}-\left(\frac{76}{28}+\frac{147}{28}\right)\)
\(=\frac{45}{4}-\frac{223}{28}\)
\(=\frac{315}{28}-\frac{223}{28}\)
\(=\frac{23}{7}\)
b) \(\left(8\frac{5}{11}+3\frac{5}{8}\right)-3\frac{5}{11}\)
\(=\left(\frac{93}{11}+\frac{29}{8}\right)-\frac{38}{11}\)
\(=\left(\frac{744}{88}+\frac{319}{88}\right)-\frac{38}{11}\)
\(=\frac{1063}{88}-\frac{38}{11}=\frac{1063}{88}-\frac{304}{88}\)
\(=\frac{69}{8}\)
a, Ta có : \(\overline{aaa}=a.111=a.3.37\Rightarrow\overline{aaa}⋮37\)
b,Vì : \(\overline{aaaaaa}=a.111111=a.15873.7\Rightarrow\overline{aaaaaa}⋮7\)
c,Vì : \(\overline{abcabc}=\overline{abc}.1001\Rightarrow\overline{abcabc}⋮1001\)
d, Ta có : \(\overline{ab}+\overline{ba}=10a+b+10b+a\)
\(=10a+a+10b+b=11a+11b\)
\(=11\left(a+b\right)⋮11\) ( Vì : \(a+b\in N\) )
Vậy \(\overline{ab}+\overline{ba}⋮11\)
e, \(\overline{ab}-\overline{ba}=\left(10a+b\right)-\left(10b+a\right)\)
\(=\left(10-1\right)a-\left(10-1\right)b\)
\(=9a-9b=9\left(a-b\right)\)
Vì : \(a\ge b\Rightarrow a-b\in N\Rightarrow9\left(a-b\right)⋮9\)
Vậy : \(\overline{ab}-\overline{ba}⋮9\)
f, \(\overline{abc}-\overline{cba}=\left(a.100+b10+c\right)-\left(100c+10b+a\right)\)
\(=\left(100a+10a+10c+c\right)-\left(100c+10c+10a+a\right)\)
\(=\left(110a+11c\right)-\left(110c+11a\right)⋮11\)
Vì : \(a\ge c\Rightarrow\overline{abc}-\overline{cba}⋮11\)
Vậy : \(\overline{abc}-\overline{cba}⋮11\)
a) \(\overline{aaa}=a.111⋮37\)
\(\Rightarrow\overline{aaa}⋮37\left(đpcm\right)\)
b) \(\overline{aaaaaa}=a.111111⋮7\) ( vì \(111111⋮7\) )
\(\Rightarrow\overline{aaaaaa}⋮7\left(đpcm\right)\)
c) \(\overline{abcabc}=\overline{abc}.1001⋮1001\)
\(\Rightarrow\overline{abcabc}⋮1001\left(đpcm\right)\)
d) \(\overline{ab}+\overline{ba}=10a+b+10b+a=11a+11b=11\left(a+b\right)⋮11\)
\(\Rightarrow\overline{ab}+\overline{ba}⋮11\left(đpcm\right)\)
e) \(\overline{ab}-\overline{ba}=10a+b-\left(10b+a\right)=9a-9b=9\left(a-b\right)⋮9\)
\(\Rightarrow\overline{ab}-\overline{ba}⋮9\left(đpcm\right)\)
f) \(\overline{abc}-\overline{cba}=100a+10b+c-100c-10b-a=99a-99c=11\left(9a-9b\right)⋮11\)
\(\Rightarrow\overline{abc}-\overline{cba}⋮11\left(đpcm\right)\)