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a)\(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{23.27}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}=\frac{1}{3}-\frac{1}{27}=\frac{8}{27}\)
b)\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)
c)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{3}-\frac{1}{13}+2\left(1-\frac{1}{10}\right)=\frac{10}{39}+\frac{9}{5}=\frac{401}{195}\)
a)\(\frac{4}{5}-\frac{1}{4}+\frac{3}{10}\)
\(=\frac{16}{20}-\frac{5}{20}+\frac{6}{20}\)
\(=\frac{17}{20}\)
b) \(\frac{2}{5}:\left(1-\frac{1}{10}\right)\)
\(=\frac{2}{5}:\frac{9}{10}\)
\(=\frac{4}{9}\)
c)\(\frac{7}{8}\times\frac{4}{9}+\frac{1}{14}:\frac{5}{14}\)
\(=\frac{7}{18}+\frac{1}{5}\)
\(=\frac{53}{90}\)
d)\(\frac{2}{7}\times\frac{3}{11}+\frac{2}{7}\times\frac{8}{11}\)
\(=\frac{2}{7}\times\left(\frac{3}{11}+\frac{8}{11}\right)\)
\(=\frac{2}{7}\times1=\frac{2}{7}\)
e) \(12+\left(16-11\right)\times4\)
\(=12+20=32\)
f)\(2\frac{3}{7}+1\frac{4}{7}\)
\(=\frac{17}{7}+\frac{11}{7}\)
\(=4\)
g)\(\frac{2}{3}\times\frac{4}{5}+\frac{1}{5}:\frac{9}{11}\)
\(=\frac{8}{15}+\frac{11}{45}\)
\(=\frac{7}{9}\)
h)\(\left(6,2:2+3,7\right):0,2\)
\(=\left(3,1+3,7\right):0,2\)
\(=6,8:0,2=34\)
#H
\(\frac{3}{5}\times\frac{2}{7}:\frac{4}{9}=\frac{3\times2}{5\times7}\times\frac{9}{4}=\frac{6}{35}\times\frac{9}{4}=\frac{54}{140}=\frac{27}{70}\)
\(\frac{2}{11}:\frac{1}{3}\times\frac{3}{2}=\frac{2}{11}\times3\times\frac{3}{2}=\frac{2\times3\times3}{11\times2}=\frac{3\times3}{11}=\frac{9}{11}\)
\(\frac{5}{2}\times\frac{1}{3}+\frac{1}{4}=\frac{5}{6}+\frac{1}{4}=\frac{10}{12}+\frac{3}{12}=\frac{13}{12}\)
\(\frac{1}{2}+\frac{1}{4}:\frac{1}{6}=\frac{1}{2}+\frac{1}{4}\times6=\frac{1}{2}+\frac{6}{4}=\frac{1}{2}+\frac{3}{2}=\frac{4}{2}=2\)
Ta có:
\(A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\)
\(A=\frac{1}{3}-\frac{1}{15}=\frac{4}{15}\)
\(B=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)\)
\(B=2.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(B=2.\left(\frac{1}{1}-\frac{1}{10}\right)=2.\frac{9}{10}\)
\(B=\frac{9}{5}\)
a) 1+2+3+...+2018
=[(2018-1):1+1].(2018+1):2
=2018.2019:2
=2037171
b) 4+7+10+13+...+2017
=[(2017-4):3+1].(2017+4):2
=672.2021:2
=679056
c) 1+2-3-4+5+6-7-8+9+10-11-12+13+14
=(1+2-3-4)+(5+6-7-8)+(9+10-11-12)+13+14
=-4+(-4)+(-4)+13+14
=-12+27
=15
d) 23 x 75 + 25 x 23 + 180
=23 x (75+25) +180
=23 x 100 +180
=2300+180
2480
chúc ban hoc tốt nha
a) Số số hạng: ( 2018 - 1 ) : 1 + 1 = 2018
Tổng: 2018 x ( 2018 + 1 ) : 2 = 2 037 171
A = ( 6 : 3/5 - 7/6 * 6/7 ) : ( 21/5 * 10/11 + 57/11 )
A = ( 10 - 1 ) : ( 42/11 + 57/11)
A = 9 : 9
A = 1
B = 59 /10 : 3/2 - ( 7/3 * 9/2 - 2 * 7/3 ) : 7/4
B = 59/15 - ( 21/2 - 14/3 ) : 7/4
B = 59/15 - 35/6 : 7/4
B = 59/15 - 10/3
B = 3/5
a) 563 x 23 + 23 x 28 - 23
= 23 x (563 + 28 - 1)
= 23 x 600
= 19200
b) \(\frac{7}{11}+\frac{3}{4}+\frac{4}{11}+\frac{1}{4}-1\frac{2}{5}=\left(\frac{7}{11}+\frac{4}{11}\right)+\left(\frac{3}{4}+\frac{1}{4}\right)-\frac{7}{5}=2-\frac{7}{5}=\frac{3}{5}\)
c) \(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times...\times\left(1-\frac{1}{10}\right)=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times....\times\frac{9}{10}\)
\(=\frac{1\times2\times3\times...\times9}{2\times3\times4\times...\times10}=\frac{1}{10}\)
d) \(\frac{2}{7}\times\frac{3}{9}+\frac{2}{7}\times\frac{2}{3}=\frac{2}{7}\times\left(\frac{3}{9}+\frac{2}{3}\right)=\frac{2}{7}\times1=\frac{2}{7}\)
a ) \(563\cdot23+23\cdot28-23\)
\(=23\cdot\left(563+28-1\right)\)
\(=23\cdot600\)
\(=19200\)
b ) \(\frac{7}{11}+\frac{3}{4}+\frac{4}{11}+\frac{1}{4}-1\frac{2}{5}=\left(\frac{7}{11}+\frac{4}{11}\right)+\left(\frac{3}{4}+\frac{1}{4}\right)-\frac{7}{5}=2-\frac{7}{5}=\frac{3}{5}\)
c ) \(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{10}\right)=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{9}{10}\)
\(=\frac{1\cdot2\cdot3\cdot...\cdot9}{2\cdot3\cdot4\cdot...\cdot10}=\frac{1}{10}\)
d ) \(\frac{2}{7}\cdot\frac{3}{9}+\frac{2}{7}\cdot\frac{2}{3}=\frac{2}{7}\cdot\left(\frac{3}{9}+\frac{2}{3}\right)=\frac{2}{7}\cdot1=\frac{2}{7}\)