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Sửa đề: bỏ 1/30*40
\(A=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{19}+...+\dfrac{1}{28}-\dfrac{1}{39}\)
=1/5-1/39
=34/195
Còn B là biểu thức nào vậy bạn?
Sửa đề: 39*40
\(A=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{12}+...+\dfrac{1}{39}-\dfrac{1}{40}=\dfrac{1}{5}-\dfrac{1}{40}=\dfrac{7}{40}\)
\(B=\dfrac{2}{3}\left(\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{17\cdot20}\right)\)
\(=\dfrac{2}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{17}-\dfrac{1}{20}\right)\)
=2/3*3/20=2/20=1/10=4/40<A
2. \(\frac{1995.1994-1}{1993.1995+1994}=\frac{1995.\left(1993+1\right)-1}{1993.1995+1994}=\frac{1995.1993+1995-1}{1993.1995+1994}=\frac{1995.1993+1994}{1993.1995+1994}\)
1. \(\frac{4}{3.7}+\frac{5}{7.12}+\frac{1}{12.13}+\frac{7}{13.20}+\frac{3}{20.23}\)
\(=\frac{7-3}{3.7}+\frac{12-7}{7.12}+\frac{13-12}{12.13}+\frac{23-20}{20.23}\)
\(=\left[\frac{7}{3.7}-\frac{3}{3.7}\right]+\left[\frac{12}{7.12}-\frac{7}{7.12}\right]+\left[\frac{13}{12.13}-\frac{12}{12.13}\right]+\left[\frac{20}{13.20}-\frac{13}{13.20}\right]+\left[\frac{23}{20.23}-\frac{20}{20.23}\right]\) \(=\left[\frac{1}{3}-\frac{1}{7}\right]+\left[\frac{1}{7}-\frac{1}{12}\right]+\left[\frac{1}{12}-\frac{1}{13}\right]+\left[\frac{1}{13}-\frac{1}{20}\right]+\left[\frac{1}{20}-\frac{1}{23}\right]\) \(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{20}+\frac{1}{20}-\frac{1}{23}\) \(=\frac{1}{3}-\frac{1}{23}\\ =\frac{20}{69}\)
tham khảo
a. 5/8
b. 29/28
c. -1/36
d. -41/24
e. -341/273
f. -17/5
g. -4
`5/39xx(7 4/5xx1 2/3+8 1/3xx7 4/5)`
`=5/39xx(39/5xx5/3+25/3xx39/5)`
`=5/39xx39/5(5/3+25/3)`
`=1xx30/3`
`=10`
a. = 125 + 170 + 1100 - 125 - 864 - 36 = 370
b. = 72 ( 7 + 3 ) + 52 ( 9 + 1 )
= 49 x 10 + 25 x 10
= 10 ( 49+25)
= 10 x 74 = 740
a.125+170+1100+(-125)+(-864)+(-36)
=1395+(-125)+(-864)+(-36)
=370
b.[(73+3.72)+(32.52+52)
=(343+3.49)+(9.25+25)
=(343+147)+(225+25)
=490+250
=740
\(A=\frac{10}{7.12}+\frac{10}{12.17}+\frac{10}{17.22}+...+\frac{10}{502.507}\)
\(=2\left(\frac{5}{7.12}+\frac{5}{12.17}+\frac{5}{17.22}+...+\frac{5}{502.507}\right)\)
\(=2\left(\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{22}+...+\frac{1}{502}-\frac{1}{507}\right)\)
\(=2\left(\frac{1}{7}-\frac{1}{507}\right)\)
\(=tự tính\)
vì @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
Rút gọn bằng kiểu nào?
\(P=\frac{5}{3\cdot7}+\frac{5}{7\cdot11}+\frac{5}{11\cdot15}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}\)
\(P=\frac{5}{4}\left(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{\left(4n-1\right)\left(4n+3\right)}\right)\)
\(P=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{4n+3}\right)\)
\(P=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{4n+3}\right)\)
...
P=\(\frac{5}{3x7}\) +\(\frac{5}{7x11}\)+\(\frac{5}{11x15}\)+...+\(\frac{5}{\left(4n-1\right)x\left(4n+3\right)}\)
\(\frac{4}{5}\)P=\(\frac{4}{3x7}\)+\(\frac{4}{7x11}\)+\(\frac{4}{11x15}\)+...+\(\frac{4}{\left(4n-1\right)x\left(4n+3\right)}\)
\(\frac{4}{5}\)P=\(\frac{1}{3}\)-\(\frac{1}{7}\)+\(\frac{1}{7}\)-\(\frac{1}{11}\)+...+\(\frac{1}{4n-1}\)-\(\frac{1}{4n+3}\)
\(\frac{4}{5}\)P=\(\frac{1}{3}\)-\(\frac{1}{4n+3}\)
P=\(\frac{5}{12}\)-\(\frac{5}{16n+12}\)
\(A=\dfrac{10}{3\cdot7}-\dfrac{1}{7}+\dfrac{1}{12}-\dfrac{1}{12}+\dfrac{1}{19}-\dfrac{1}{19}+\dfrac{1}{24}=\dfrac{10}{21}+\dfrac{1}{24}=\dfrac{29}{56}\)
bn nên gõ latex để mn dễ hỗ trợ trl nha