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31 tháng 7 2015
A = 5/3.7 + 5/7.11 + 5/11.15 + ... + 5/35.39
A = 5/4 x ( 4/3.7 + 4/7.11 + 4/11.15 + ... + 4/35.39 )
A = 5/4 x ( 1/3 - 1/7 + 1/7 - 1/11 + 1/11 - 1/15 + ... + 1/35 - 1/39 )
A = 5/4 x ( 1/3 - 1/39 )
A = 5/4 x 4/13
A = 5/13
NQ
1
14 tháng 4 2023
a: =35/17-18/17-9/5+4/5
=1-1=0
b: =-7/19(3/17+8/11-1)
=7/19*18/187=126/3553
c: =26/15-11/15-17/3-6/13
=1-6/13-17/3
=7/13-17/3=-200/39
MN
0
5 tháng 3 2023
a: \(=\left(15-6-\dfrac{13}{18}\right):\dfrac{298}{27}-\dfrac{17}{8}:\dfrac{51}{40}\)
\(=\dfrac{149}{18}\cdot\dfrac{27}{298}-\dfrac{5}{3}=\dfrac{3}{2}-\dfrac{5}{3}=\dfrac{9-10}{6}=\dfrac{-1}{6}\)
b: \(=\dfrac{-16}{5}\cdot\dfrac{-15}{64}+\dfrac{-22}{15}:\dfrac{11}{2}\)
\(=\dfrac{3}{4}-\dfrac{4}{15}=\dfrac{29}{60}\)
c: \(=\dfrac{-7}{9}\left(\dfrac{4}{11}+\dfrac{7}{11}\right)+5+\dfrac{7}{9}=\dfrac{-7}{9}+\dfrac{7}{9}+5=5\)
d: \(=\dfrac{1}{2}\cdot\dfrac{4}{3}\cdot10\cdot\dfrac{1}{5}\cdot\dfrac{3}{4}=1\)
e: \(=\dfrac{4}{25}+\dfrac{11}{2}\cdot\dfrac{5}{2}+\dfrac{-23}{4}=\dfrac{204}{25}\)
2 tháng 5 2019
a) \(\frac{8}{40}+\frac{-14}{35}-\frac{12}{60}\)
= \(\frac{1}{5}-\frac{2}{5}-\frac{1}{5}\)
= \(\left(\frac{1}{5}-\frac{1}{5}\right)-\frac{2}{5}\)
= \(-\frac{2}{5}\)
b) 5/7.5/11 + 5/7.5/11 - 5/7.14/11
= 5/7.(5/11 + 5/11 - 14/11)
= 5/7.(-4/11)
= -20/77
c) \(19\frac{5}{8}:\frac{7}{12}-15\frac{1}{4}:\frac{7}{12}\)
= \(\left(19\frac{5}{8}-15\frac{1}{4}\right):\frac{7}{12}\)
= \(\frac{35}{8}:\frac{7}{12}\)
= \(\frac{15}{2}\)
2 tháng 5 2019
d) 2/5.1/3 - 2/15 : 1/5 + 3/5.1/3
= (2/5 + 3/5).1/3 - 2/15 . 5
= 1.1/3 - 2/3
= 1/3 - 2/3
= -1/3
e) \(\frac{4}{9}.19\frac{1}{3}-\frac{4}{9}.39\frac{1}{3}\)
= \(\frac{4}{9}.\left(19\frac{1}{3}-39\frac{1}{3}\right)\)
= \(\frac{4}{9}.\left(-20\right)\)
= \(\frac{-80}{9}\)
g) \(\frac{81.17-15.81}{81.17-81.15}\)
= 1
\(A=\frac{5}{3\cdot7}+\frac{5}{7\cdot11}+\frac{5}{11\cdot15}+...+\frac{5}{35\cdot39}\)
\(=\frac{5}{4}\cdot\left(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{35\cdot39}\right)\)
\(=\frac{5}{4}\cdot\left(1-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...-\frac{1}{35}+\frac{1}{35}-\frac{1}{39}\right)\)
\(=\frac{5}{4}\cdot\left(1-\frac{1}{39}\right)=\frac{5}{4}\cdot\frac{38}{39}=\frac{95}{78}\)