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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
3 tháng 7
c; 17\(\dfrac{2}{31}\) - (\(\dfrac{15}{17}\) + 6\(\dfrac{2}{31}\))
= 17 + \(\dfrac{2}{31}\) - \(\dfrac{15}{17}\) - 6 - \(\dfrac{2}{31}\)
= (17 - 6) - \(\dfrac{15}{17}\) + (\(\dfrac{2}{31}\) - \(\dfrac{2}{31}\))
= 11 - \(\dfrac{15}{17}\)+ 0
= \(\dfrac{172}{17}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
3 tháng 7
b; 130\(\dfrac{25}{28}\) + 120\(\dfrac{17}{35}\)
= 130 + \(\dfrac{25}{28}\) + 120 + \(\dfrac{17}{35}\)
= (130 + 120) + (\(\dfrac{25}{28}\) + \(\dfrac{17}{35}\))
= 250 + (\(\dfrac{125}{140}\) + \(\dfrac{68}{140}\))
= 250 + \(\dfrac{193}{140}\)
= 250\(\dfrac{193}{140}\)
CM
1
14 tháng 8 2023
c: C=1*2+2*3+3*4+...+58*59+59*60
=>3*C=1*2*3+2*3*(4-1)+3*4*(5-2)+...+58*59(60-57)+59*60(61-58)
=>3*C=1*2*3+2*3*4-1*2*3+...+58*59*60-58*59*57+59*60*61-58*59*60
=>3*C=59*60*61
=>C=59*20*61=71980
DP
31 tháng 7 2017
\(\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+...+\frac{5}{x\left(x+1\right)}=\frac{64}{13}\)
\(\Leftrightarrow5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{64}{13}\)
\(\Leftrightarrow1-\frac{1}{x+1}=\frac{64}{13}\div5\)
\(\Leftrightarrow1-\frac{1}{x+1}=\frac{64}{65}\)
\(\Leftrightarrow\frac{1}{x+1}=1-\frac{64}{65}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{65}\)
\(\Rightarrow x+1=65\Rightarrow x=65-1=64\)
\(\text{Vậy }x=64\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
19 tháng 8 2023
c \(\dfrac{2}{3}\times\dfrac{13}{4}\times\dfrac{3}{2}\times\dfrac{7}{3}\times\dfrac{4}{7}\)
= \(\dfrac{2\times13\times3\times7\times4}{3\times4\times2\times3\times7}\)
= \(\dfrac{13}{3}\)
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}\)
=\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}\)
=\(\dfrac{1}{1}-\dfrac{1}{5}\)
=\(\dfrac{4}{5}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
19 tháng 8 2023
a, (\(\dfrac{41}{25}\) + \(\dfrac{17}{100}\)) + \(\dfrac{39}{25}\)
= (\(\dfrac{41}{25}\) + \(\dfrac{39}{25}\)) + \(\dfrac{17}{100}\)
= \(\dfrac{80}{25}\) + \(\dfrac{17}{100}\)
= \(\dfrac{320}{100}\) + \(\dfrac{17}{100}\)
= \(\dfrac{337}{100}\)
g, \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\) + \(\dfrac{1}{30}\) + \(\dfrac{1}{42}\)
= \(\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}\)
= \(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\)
= \(\dfrac{1}{3}-\dfrac{1}{7}\)
= \(\dfrac{7}{21}-\dfrac{3}{21}\)
= \(\dfrac{4}{21}\)
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
3 tháng 8 2018
A = 1 + 2 + 3 + ... + 2018 (có 2018 số )
= (2018 + 1) . 2018 : 2 = 2037171
B = 1 + 3 + 5 + ... + 2017(có 1009 số )
= (2017 + 1) . 1009 : 2 = 1018081
C = 2 + 4 + 6 + ... + 2018 (Có 1009 số )
= (2018 + 2) x 1009 : 2 = 1019090
D = 72 . 153 + 27.153 + 153
= (72 + 27 + 1) . 153
= 100 . 153 = 15300
NT
0
Phần A sai nha phải là: 10 + 13 + ... + 79 + 82
đáp án:
SSH: A = ( 82 - 10 ) : 3 + 1 = 25
Tổng: ( 10 + 82 ) . 25 : 2 = 1150
a) A = 10+13 + ...+79 + 81
A = ( 79+10) x 24 : 2 + 81
A = 89 x 24 : 2 +81
A = 1068+ 81
A= 1149
b) chỗ " 1/12.20" phải là 1/12.13 chứ !
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{12.13}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{12}-\frac{1}{13}\)
\(=1-\frac{1}{13}\)
\(=\frac{12}{13}\)
c) \(C=\frac{3}{7}\times\frac{4}{13}+\frac{3}{7}+5\frac{4}{7}\)
\(C=\frac{3}{7}\times\left(\frac{4}{13}+1\right)+\frac{39}{7}\)
\(C=\frac{3}{7}\times\frac{17}{13}+\frac{39}{7}\)
\(C=\frac{51}{91}+\frac{39}{7}\)
\(C=\frac{558}{91}\)