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\(\frac{2.4}{6.18}=\frac{1.2}{3.9}=\frac{2}{27}\)
\(\frac{3.5.7}{6.9.14}=\frac{1.5.7}{2.9.2}=\frac{35}{36}\)
\(\frac{4.7-4.5}{64}=\frac{4\left(7-5\right)}{64}=\frac{8}{64}=\frac{1}{8}\)
a)\(\frac{2.4}{6.18}\)=\(\frac{2.2.2}{2.3.2.3.3}\)=\(\frac{2.1.1}{1.3.1.3.3}\)=\(\frac{2}{27}\)
b)\(\frac{3.5.7}{6.9.14}\)=\(\frac{3.5.7}{2.3.3.3.2.7}\)=\(\frac{1.5.1}{2.1.3.3.2.1}\)=\(\frac{5}{36}\)
c)\(\frac{4.7-4.5}{64}\)=\(\frac{4.\left(7-5\right)}{64}\)=\(\frac{4.2}{64}\)=\(\frac{8}{64}\)=\(\frac{1}{8}\)
a: \(\dfrac{4\cdot5+4\cdot11}{8\cdot7+4\cdot3}=\dfrac{20+44}{56+12}=\dfrac{64}{68}=\dfrac{16}{17}=\dfrac{11088}{11781}\)
\(\dfrac{-15\cdot8+10\cdot7}{5\cdot6+20\cdot3}=\dfrac{-120+70}{30+60}=\dfrac{-50}{90}=\dfrac{-5}{9}=\dfrac{-6545}{11781}\)
\(\dfrac{2^4\cdot5^2\cdot7}{2^3\cdot5\cdot7^2\cdot11}=\dfrac{2\cdot5}{7\cdot11}=\dfrac{10}{77}=\dfrac{1530}{11781}\)
Câu 1:
\(\dfrac{14\cdot34-21\cdot10}{28\cdot10-28\cdot16}=\dfrac{7\cdot2\cdot34-7\cdot3\cdot10}{28\left(10-16\right)}\)
\(=\dfrac{7\left(2\cdot34-3\cdot10\right)}{28\cdot\left(-6\right)}=\dfrac{1}{4}\cdot\dfrac{68-30}{-6}=\dfrac{1}{4}\cdot\dfrac{38}{-6}=\dfrac{38}{-24}=-\dfrac{19}{12}\)
a, A = 1 - 1/2 + 1/2 - 1/3 + 1/3 -1/4 +... + 1/2017 - 1/2018
A = 1 - 1/2018 = 2017/2018
b, B = 5/2 . ( 1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 + ... + 1/2016 -1/2018)
B= 5/2 . ( 1/2 - 1/ 2018 )
B = 504/1009
c, C = 1/3.6 + 1/ 6.9 + 1/ 9.12 + ... + 1/ 30.33
C= 1/3 - 1/6 + 1/6 - 1/ 9 + 1/9 - 1/12 + ... + 1/30 - 1/33
C = 1/3 - 1/33
C= 10/33
phan B mk quên nhân với 5/2
lấy 5/2 . 504/1009 = 1260/1009
S = \(\dfrac{1}{1.4}\)+ \(\dfrac{1}{4.7}\)+...+\(\dfrac{1}{2002.2005}\)
S = ( 1 - \(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-...+\(\dfrac{1}{2002}\)-\(\dfrac{1}{2005}\)) . \(\dfrac{1}{3}\)
S = ( 1 - \(\dfrac{1}{2005}\)) . \(\dfrac{1}{3}\)
S = \(\dfrac{2004}{2005}\). \(\dfrac{1}{3}\)
S = \(\dfrac{2014}{6015}\)
a) \(S=\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{2002.2005}\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{2002}-\dfrac{1}{2005}\right)\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{2005}\right)\)
\(=\dfrac{1}{3}.\dfrac{2004}{2005}=\dfrac{668}{2005}\)
KL.
b) \(P=\dfrac{3}{1.6}+\dfrac{3}{6.11}+\dfrac{3}{11.16}+...+\dfrac{3}{96.101}\)
\(=\dfrac{3}{5}\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{96}-\dfrac{1}{101}\right)\)
\(=\dfrac{3}{5}\left(1-\dfrac{1}{101}\right)\)
\(=\dfrac{3}{5}.\dfrac{100}{101}=\dfrac{60}{101}\)
KL.
c) \(Q=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{98.99.100}\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{99.100}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{9900}\right)\)
\(=\dfrac{1}{2}.\dfrac{1}{19800}=\dfrac{1}{39600}\)
KL.
Đặt \(A=\dfrac{1}{1.2}+\dfrac{2}{2.4}+\dfrac{3}{4.7}+\dfrac{4}{7.11}+\dfrac{5}{11.16}+\dfrac{6}{16.22}\)
\(1A=1-\left(\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(\dfrac{1}{7}+\dfrac{1}{7}\right)+\left(\dfrac{1}{11}+\dfrac{1}{11}\right)+\left(\dfrac{1}{16}+\dfrac{1}{16}\right)-\dfrac{1}{22}\)\(1A=1-\dfrac{1}{22}\)
\(1A=\dfrac{22}{22}-\dfrac{1}{22}\)
\(1A=\dfrac{21}{22}\)
\(\dfrac{21}{22}\) không thể rút gọn
\(A=\dfrac{1}{1\cdot2}+\dfrac{2}{2\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{5}{11\cdot16}+\dfrac{6}{16\cdot22}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{22}\\ =1-\dfrac{1}{22}\\ =\dfrac{21}{22}\)
Vậy \(A=\dfrac{21}{22}\)
A = \(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
A=\(\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{50}{100}-\dfrac{1}{100}=\dfrac{49}{100}\)
B = \(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{49.51}\)
B = \(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{49}-\dfrac{1}{51}\)
B = \(\dfrac{1}{2}-\dfrac{1}{51}=\dfrac{51}{102}-\dfrac{2}{102}=\dfrac{49}{102}\)
a, \(\dfrac{2.4}{6.18}\)=\(\dfrac{2}{27}\)
b,\(\dfrac{3.5.7}{6.9.14}\)=\(\dfrac{105}{756}\)=\(\dfrac{35}{232}\)
c,\(\dfrac{4.7-4.5}{64}\)=\(\dfrac{28-20}{64}\)=\(\dfrac{8}{64}\)=\(\dfrac{1}{8}\)
nhớ like nhé
\(\dfrac{2.4}{6.18}=\dfrac{8}{108}=\dfrac{2}{27}\)
\(\dfrac{3.5.7}{6.9.14}=\dfrac{105}{756}=\dfrac{35}{252}\)
\(\dfrac{4.7-4.5}{64}=\dfrac{4.\left(7-5\right)}{64}=\dfrac{4.2}{64}=\dfrac{8}{64}=\dfrac{1}{8}\)
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