NV
1
K
Khách
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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
VT
20 tháng 8 2017
\(a,=\frac{7-1}{1.3.7}+\frac{9-3}{3.7.9}+\frac{13-7}{7.9.13}+\frac{15-9}{9.13.15}\)\(+\frac{19-13}{13.15.19}\)
\(=\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+\frac{1}{7.9}-\frac{1}{9.13}+\frac{1}{9.13}-\frac{1}{13.15}+\frac{1}{13.15}-\frac{1}{15.19}\)
\(=\frac{1}{1.3}-\frac{1}{15.19}=\frac{95}{285}-\frac{1}{285}=\frac{94}{285}\)
\(b,=\frac{1}{6}.\left(\frac{6}{1.3.7}+\frac{6}{3.7.9}+\frac{6}{7.9.13}+\frac{6}{9.13.15}+\frac{6}{13.15.19}\right)\)
làm giống như trên
\(c,=\frac{1}{8}.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{48.49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{50-48}{48.49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{1}{2}-\frac{1}{2450}\right)=\frac{1}{16}.\left(\frac{1225}{2450}-\frac{1}{2450}\right)=\frac{153}{4900}\)
\(d,=\frac{5}{7}.\left(\frac{7}{1.5.8}+\frac{7}{5.8.12}+\frac{7}{8.12.15}+...+\frac{7}{33.36.40}\right)\)
\(=\frac{5}{7}.\left(\frac{8-1}{1.5.8}+\frac{12-5}{5.8.12}+\frac{15-8}{8.12.15}+...+\frac{40-33}{33.36.40}\right)\)
\(=\frac{5}{7}.\left(\frac{1}{1.5}-\frac{1}{5.8}+\frac{1}{5.8}-\frac{1}{8.12}+\frac{1}{8.12}-\frac{1}{12.15}+...+\frac{1}{33.36}-\frac{1}{36.40}\right)\)
\(=\frac{5}{7}.\left(\frac{1}{5}-\frac{1}{1440}\right)=\frac{5}{7}.\left(\frac{288}{1440}-\frac{1}{1440}\right)=\frac{41}{288}\)
P/S: . là nhân nha
ND
1 tháng 11 2020
cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc
1 tháng 11 2020
\(\frac{4}{6}+\frac{7}{13}+\frac{17}{9}+\frac{19}{13}-\frac{8}{9}+\frac{1}{6}\)
\(=\left(\frac{4}{6}+\frac{1}{6}\right)+\left(\frac{7}{13}+\frac{19}{13}\right)+\left(\frac{17}{9}-\frac{8}{9}\right)\)
\(=\frac{5}{6}+2+1\)
\(=3\frac{5}{6}=\frac{7}{2}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
5 tháng 8 2023
a, \(\dfrac{7}{8}\) \(\times\) \(\dfrac{3}{13}\) + \(\dfrac{4}{9}\) \(\times\) \(\dfrac{4}{13}\)
= \(\dfrac{1}{13}\) \(\times\)( \(\dfrac{21}{8}\) + \(\dfrac{16}{9}\))
= \(\dfrac{1}{13}\) \(\times\)( \(\dfrac{189}{72}\) + \(\dfrac{128}{72}\))
= \(\dfrac{1}{13}\) \(\times\) \(\dfrac{317}{73}\)
= \(\dfrac{317}{949}\)
b, \(\dfrac{6}{5}\) + \(\dfrac{7}{3}\) + \(\dfrac{8}{9}\)
= \(\dfrac{54}{45}\) + \(\dfrac{105}{45}\) + \(\dfrac{40}{45}\)
= \(\dfrac{199}{45}\)
c, 23 : \(\dfrac{5}{14}\) + \(\dfrac{6}{7}\) + \(\dfrac{4}{9}\)
= \(\dfrac{322}{5}\) + \(\dfrac{6}{7}\) + \(\dfrac{4}{9}\)
= \(\dfrac{20286}{315}\) + \(\dfrac{270}{315}\) + \(\dfrac{140}{315}\)
= \(\dfrac{20696}{315}\)
d, 4\(\dfrac{1}{4}\) + 7\(\dfrac{3}{7}\) - 2\(\dfrac{4}{17}\)
= 4 + \(\dfrac{1}{4}\) + 7 + \(\dfrac{3}{7}\) - 2 - \(\dfrac{4}{17}\)
= (4+7-2) + (\(\dfrac{1}{4}\) + \(\dfrac{3}{7}\) - \(\dfrac{4}{17}\))
= 9 + \(\dfrac{119}{476}\) + \(\dfrac{204}{476}\) - \(\dfrac{112}{476}\)
= 9\(\dfrac{211}{476}\) = \(\dfrac{4495}{476}\)
e, 8 - (9\(\dfrac{2}{11}\) + \(\dfrac{8}{33}\))
= 8 - 9 - \(\dfrac{2}{11}\) - \(\dfrac{8}{33}\)
= -1 - \(\dfrac{2}{11}\) - \(\dfrac{8}{33}\)
= \(\dfrac{-33}{33}\) - \(\dfrac{-6}{33}\) - \(\dfrac{8}{33}\)
= - \(\dfrac{47}{33}\)
C
1
H9
HT.Phong (9A5)
CTVHS
24 tháng 6 2023
a) \(\dfrac{4}{6}+\dfrac{7}{13}+\dfrac{17}{9}-\dfrac{8}{9}+\dfrac{14}{6}+\dfrac{19}{13}\)
\(=\dfrac{2}{3}+\dfrac{7}{13}+\dfrac{17}{9}-\dfrac{8}{9}+\dfrac{7}{3}+\dfrac{19}{13}\)
\(=\left(\dfrac{2}{3}+\dfrac{7}{3}\right)+\left(\dfrac{7}{13}+\dfrac{19}{13}\right)+\left(\dfrac{17}{9}-\dfrac{8}{9}\right)\)
\(=\dfrac{9}{3}+\dfrac{26}{13}+\dfrac{9}{9}\)
\(=3+2+1\)
\(=6\)
b) \(\dfrac{328}{435}\times\dfrac{468}{432}\times\dfrac{435}{164}\times\dfrac{432}{984}\times\dfrac{164}{468}\)
\(=\dfrac{328\times468\times435\times432\times164}{435\times432\times164\times984\times468}\)
\(=\dfrac{328}{984}\)
\(=\dfrac{1}{3}\)
NT
2
ND
Nguyễn Đức Trí
VIP
28 tháng 7 2023
Tử số chung là \(2^3.3^2.11=792\)
\(\dfrac{11}{13}=\dfrac{11x72}{13x72}=\dfrac{792}{936}\)
\(\dfrac{6}{8}=\dfrac{6x12x11}{8x12x11}=\dfrac{792}{1056}\)
\(\dfrac{4}{5}=\dfrac{4x18x11}{5x18x11}=\dfrac{792}{990}\)
\(\dfrac{8}{9}=\dfrac{8x9x11}{9x9x11}=\dfrac{792}{891}\)
\(\dfrac{9}{11}=\dfrac{9x8x11}{11x8x11}=\dfrac{792}{968}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
28 tháng 7 2023
11 = 1 \(\times\) 11; 6 = 2 \(\times\) 3; 8 = 2 \(\times\) 2 \(\times\) 2; 9 = 3 \(\times\) 3
Vậy tử số chung là: 11 \(\times\) 2 \(\times\)2 \(\times\) 2 \(\times\) 3 \(\times\) 3 = 792
\(\dfrac{11}{13}\) = \(\dfrac{11\times72}{13\times72}\) = \(\dfrac{792}{936}\)
\(\dfrac{6}{8}\) = \(\dfrac{6\times132}{8\times132}\) = \(\dfrac{792}{1056}\)
\(\dfrac{4}{5}\) = \(\dfrac{4\times198}{5\times198}\) = \(\dfrac{792}{990}\)
\(\dfrac{8}{9}\) = \(\dfrac{8\times99}{9\times99}\) = \(\dfrac{792}{891}\)
\(\dfrac{9}{11}\) = \(\dfrac{9\times88}{11\times88}\) = \(\dfrac{792}{968}\)
CB
1
ZV
0
\(\dfrac{6}{9}< \dfrac{13}{x}< \) \(\dfrac{6}{8}\)
\(\dfrac{2}{3}\) < \(\dfrac{13}{x}\) < \(\dfrac{3}{4}\)
\(\dfrac{78}{117}\) < \(\dfrac{78}{6x}\) < \(\dfrac{78}{104}\)
117 > 6\(x\) > 104
117: 6 > \(x\) > 104: 6
19,5 > \(x\) > 17,(3)
\(x\) là số tự nhiên nên \(x\) = 18