A=(1+1/3)(1+1/8)(1+1/15)...(1+1/2400)
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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
19 tháng 5 2023
a, \(\dfrac{10}{17}\) + \(\dfrac{5}{-13}\) - \(\dfrac{11}{25}\) + \(\dfrac{7}{17}\) - \(\dfrac{8}{13}\)
= ( \(\dfrac{10}{17}\) + \(\dfrac{7}{17}\)) - ( \(\dfrac{5}{13}\) + \(\dfrac{8}{13}\)) - \(\dfrac{11}{25}\)
= \(\dfrac{17}{17}\) - \(\dfrac{13}{13}\) - \(\dfrac{11}{25}\)
= 1 - 1 - \(\dfrac{11}{25}\)
= - \(\dfrac{11}{25}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
19 tháng 5 2023
b, 0,3 - \(\dfrac{93}{7}\) - 70% - \(\dfrac{4}{7}\)
= 0,3 - 0,7 - ( \(\dfrac{93}{7}+\dfrac{4}{7}\))
= - 0,4 - \(\dfrac{97}{7}\)
= - \(\dfrac{2}{5}\) - \(\dfrac{97}{7}\)
= - \(\dfrac{499}{35}\)
TA
0
8 tháng 4 2021
Giải:
A=5/9+2/15-6/9
=(5/9-6/9)+2/15
= -1/9 + 2/15
= 1/45
B=2/7-3/8+4/7+1/7-5/8+5/15
= (2/7+4/7+1/7) + (-3/8-5/8) +1/3
= 1+ (-1) +1/3
=1/3
C=3/5+1/15+1/57+1/3-2/9-3/4-1/36
=9/15+1/15+1/57+19/57-8/36-27/36-1/36
=(9/15+1/15)+(1/57+19/57)+(-8/36-27/36-1/36)
=2/3+20/57+(-1)
=58/57+(-1)
=1/57
D=1/1.2+1/2.3+1/3.4+...+1/99.100
=1/1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
=1/1-1/100
=99/100
Câu E mình ko biết làm nhé!
HN
19 tháng 7 2016
a) \(\frac{15}{16}\cdot\frac{4}{3}-\frac{1}{2}:\frac{5}{4}+3\)
\(=\frac{5}{4}-\frac{2}{5}+3\)
\(=\frac{25}{20}-\frac{8}{20}+\frac{60}{20}\)
\(=\frac{77}{20}=3\frac{17}{20}\)
b) \(\left(\frac{2}{3}-\frac{3}{8}\right):\left(\frac{3}{5}+\frac{1}{4}\right)\)
\(=\frac{7}{24}:\frac{17}{20}\)
\(=\frac{35}{102}\)
c) \(\frac{15}{8}\cdot\left(\frac{1}{3}+\frac{1}{8}\right)-\frac{3}{8}:\frac{3}{4}\)
\(=\frac{15}{8}\cdot\frac{11}{24}-\frac{1}{2}\)
\(=\frac{55}{64}-\frac{1}{2}\)
\(=\frac{23}{64}\)
d) \(\frac{20}{21}:\left(\frac{4}{5}-\frac{1}{10}\right)+\frac{13}{15}:\frac{5}{26}\)
\(=\frac{20}{21}:\frac{7}{10}+\frac{52}{15}\)
\(=\frac{200}{147}+\frac{52}{15}\)
\(=4\frac{608}{735}\)
10 tháng 8 2022
1.3.77−1+3.7.99−3+7.9.1313−7+9.13.1515−9+\frac{19-13}{13.15.19}+13.15.1919−13
=\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+\frac{1}{7.9}-\frac{1}{9.13}+\frac{1}{9.13}-\frac{1}{13.15}+\frac{1}{13.15}-\frac{1}{15.19}=1.31−3.71+3.71−7.91+7.91−9.131+9.131−13.151+13.151−15.191
=\frac{1}{1.3}-\frac{1}{15.19}=\frac{95}{285}-\frac{1}{285}=\frac{94}{285}=1.31−15.191=28595−2851=28594
b,=\frac{1}{6}.\left(\frac{6}{1.3.7}+\frac{6}{3.7.9}+\frac{6}{7.9.13}+\frac{6}{9.13.15}+\frac{6}{13.15.19}\right)b,=61.(1.3.76+3.7.96+7.9.136+9.13.156+13.15.196)
làm giống như trên
c,=\frac{1}{8}.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\right)c,=81.(1.2.31+2.3.41+3.4.51+...+48.49.501)
=\frac{1}{16}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{48.49.50}\right)=161.(1.2.32+2.3.42+3.4.52+...+48.49.502)
=\frac{1}{16}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{50-48}{48.49.50}\right)=161.(1.2.33−1+2.3.44−2+3.4.55−3+...+48.49.5050−48)
=\frac{1}{16}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)=161.(1.21−2.31+2.31−3.41+3.41−4.51+...+48.491−49.501)
=\frac{1}{16}.\left(\frac{1}{2}-\frac{1}{2450}\right)=\frac{1}{16}.\left(\frac{1225}{2450}-\frac{1}{2450}\right)=\frac{153}{4900}=161.(21−24501)=161.(24501225−24501)=4900153
d,=\frac{5}{7}.\left(\frac{7}{1.5.8}+\frac{7}{5.8.12}+\frac{7}{8.12.15}+...+\frac{7}{33.36.40}\right)d,=75.(1.5.87+5.8.127+8.12.157+...+33.36.407)
=\frac{5}{7}.\left(\frac{8-1}{1.5.8}+\frac{12-5}{5.8.12}+\frac{15-8}{8.12.15}+...+\frac{40-33}{33.36.40}\right)=75.(1.5.88−1+5.8.1212−5+8.12.1515−8+...+33.36.4040−33)
=\frac{5}{7}.\left(\frac{1}{1.5}-\frac{1}{5.8}+\frac{1}{5.8}-\frac{1}{8.12}+\frac{1}{8.12}-\frac{1}{12.15}+...+\frac{1}{33.36}-\frac{1}{36.40}\right)=75.(1.51−5.81+5.81−8.121+8.121−12.151+...+33.361−36.401)
=\frac{5}{7}.\left(\frac{1}{5}-\frac{1}{1440}\right)=\frac{5}{7}.\left(\frac{288}{1440}-\frac{1}{1440}\right)=\frac{41}{288}=75.(51−14401)=75.(1440288−14401)=28841
P/S: . là nhân nha
\(A=\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{8}\right)\left(1+\dfrac{1}{15}\right)...\left(1+\dfrac{1}{2400}\right)\)
\(=\dfrac{4}{3}.\dfrac{9}{8}...\dfrac{2401}{2400}\)
\(=\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}...\dfrac{49.49}{48.50}\)
\(=\dfrac{2.3...49}{2.3...49}.\dfrac{2.3...49}{3.4...50}=1.\dfrac{2.49}{50}=\dfrac{49}{25}\)
\(A=\left(1+\dfrac{1}{3}\right)\cdot\left(1+\dfrac{1}{8}\right)\cdot\left(1+\dfrac{1}{15}\right)\cdot\cdot\cdot\left(1+\dfrac{1}{2400}\right)\)
\(A=\left(\dfrac{3}{3}+\dfrac{1}{3}\right)\cdot\left(\dfrac{8}{8}+\dfrac{1}{8}\right)\cdot\left(\dfrac{15}{15}+\dfrac{1}{15}\right)\cdot\cdot\cdot\left(\dfrac{2400}{2400}+\dfrac{1}{2400}\right)\)
\(A=\dfrac{4}{3}\cdot\dfrac{9}{8}\cdot\dfrac{16}{15}\cdot\cdot\cdot\dfrac{2401}{2400}\)
\(A=\)\(\dfrac{2\cdot2}{1\cdot3}\cdot\dfrac{3\cdot3}{2\cdot4}\cdot\dfrac{4\cdot4}{3\cdot5}\cdot\cdot\cdot\dfrac{49\cdot49}{48\cdot50}\)
\(A=\dfrac{\left(2\cdot3\cdot4\cdot49\right)}{\left(1\cdot2\cdot3\cdot48\right)}\cdot\dfrac{\left(2\cdot3\cdot4\cdot\cdot\cdot49\right)}{\left(3\cdot4\cdot5\cdot\cdot\cdot50\right)}\)
\(A=\dfrac{49\cdot2}{50}=\dfrac{98}{50}=\dfrac{49}{35}\)