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câu 1 / 2242,82 : 100 + 37411,8 : 1000 = 59,84
câu 2 / (91,5 x 26,3 - 17,4 x 25,6) x ( 41 x 11 - 4100 x 0,1 -41 ) = 0
k với nhé !!!!!!!!!!hihi!!!!!!!!
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\(5.25.2.41.8\\ =\left(25.8\right).\left(2.5\right).41\\ =200.10.41\\ =2000.41\\ =82000\)
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\(\frac{\left(\frac{36}{41}:\frac{9}{41}\right)}{\left(\frac{14}{21}:\frac{7}{21}\right)}\times\frac{2}{5}\)
\(=\frac{\left(\frac{36}{41}\times\frac{41}{9}\right)}{\left(\frac{14}{21}\times\frac{21}{7}\right)}\times\frac{2}{5}\)
\(=\frac{4}{2}\times\frac{2}{5}\)
\(=\frac{4}{5}\)
~ Ủng hộ nhé
\(\frac{\text{36/41 : 9/41 }}{\text{14/21: 7/21}}\)= \(\frac{4}{2}\)= 2 => 2 x \(\frac{2}{5}\)= \(\frac{4}{5}\)
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Bài 1 :
a) 37581 - 9999 = 37581 - 10000 - 1 = 27581 - 1 = 27580
k cho mình nha bạn !
a)37581-9999=37581-(9999+1-1)=37581-(10000+1)=37581-10000-1=27581+1=27582
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\(\left(82-41\cdot2\right):36\cdot\left(32+17+99-81+1\right)\)
\(=\left(82-82\right):36\cdot\left[\left(32+17+1\right)+99-81\right]\)
\(=0:36\cdot\left[\left(32+17+1\right)+99-81\right]\)
\(=0\cdot\left[\left(32+17+1\right)+99-81\right]\)
\(=0\)
Học tốt !!!
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\(\frac{70}{3}\left(\frac{39}{30}+\frac{39}{42}\right)-\frac{246}{7}\div\left(\frac{41}{56}+\frac{41}{72}\right)\)
\(=\frac{70}{3}\left(\frac{13}{10}+\frac{13}{14}\right)-\frac{246}{7}\div\left(\frac{41}{7\cdot8}+\frac{41}{8\cdot9}\right)\)
\(=\frac{70}{3}\left(1+\frac{3}{10}+1-\frac{1}{14}\right)-\frac{246}{7}\div\left(\frac{40+1}{7\cdot8}+\frac{40+1}{8\cdot9}\right)\)
\(=\frac{70}{3}\left[\left(1+1\right)+\left(\frac{3}{10}-\frac{1}{14}\right)\right]-\frac{246}{7}\div\left(\frac{5}{7}+\frac{1}{7\cdot8}+\frac{5}{9}+\frac{1}{8\cdot9}\right)\)
\(=\frac{70}{3}\left(2+\frac{8}{35}\right)-\frac{246}{7}\div\left[\frac{5}{7}+\frac{5}{9}+\left(\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)\right]\)
\(=\frac{70}{3}\cdot\frac{78}{35}-\frac{246}{7}\div\left[\frac{5}{7}+\frac{5}{9}+\left(\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\right]\)
\(=\frac{35\cdot2\cdot26\cdot3}{3\cdot35}-\frac{246}{7}\div\left(\frac{5}{7}+\frac{5}{9}+\frac{1}{7}-\frac{1}{9}\right)\)
\(=52-\frac{246}{7}\div\left[\left(\frac{5}{7}+\frac{1}{7}\right)+\left(\frac{5}{9}-\frac{1}{9}\right)\right]\)
\(=52-\frac{246}{7}\div\left(\frac{6}{7}+\frac{4}{9}\right)\)
\(=52-\frac{246}{7}\div\frac{82}{63}\)
\(=52-\frac{82\cdot3\cdot9\cdot7}{7\cdot82}\)
\(=52-27=25\)
\(\frac{57}{20}-\frac{26}{15}+\frac{139}{20}\div3\)
\(=\frac{57}{20}-\frac{26}{15}+\frac{139}{60}\)
\(=\frac{171}{60}-\frac{104}{60}+\frac{139}{60}=\frac{103}{30}\)
\(\frac{39}{4}+\frac{2}{3}\left(11-\frac{23}{4}\right)\)
\(=\frac{39}{4}+11\cdot\frac{2}{3}-\frac{23}{4}\cdot\frac{2}{3}\)
\(=\frac{39}{4}+\frac{22}{3}-\frac{56}{12}\)
\(=\frac{119}{12}+\frac{88}{12}-\frac{56}{12}=\frac{151}{12}\)
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2002}\right)\left(1-\frac{1}{2003}\right)\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2001}{2002}\cdot\frac{2002}{2003}\cdot\frac{2003}{2004}\)
\(=\frac{1\cdot2\cdot3\cdot...\cdot2001\cdot2002\cdot2003}{2\cdot3\cdot4\cdot...\cdot2002\cdot2003\cdot2004}=\frac{1}{2004}\)
41 : 2 = 20,5