1 1/17 . 1 1/24 . (-5,1)
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LH
1
13 tháng 7 2019
\(1\frac{1}{7}.1\frac{1}{24}.\left(-5,1\right)\)
\(=\frac{8}{7}\cdot\frac{25}{24}\cdot\left(-5,1\right)\)
\(=\frac{25}{21}.\left(-5,1\right)\)
\(=-\frac{85}{14}\)
15 tháng 10 2021
\(B=\left\{x\in Z|-2021< x< 1\right\}\\ B.có:2020+0-1=2021\left(phần.tử\right)\\ C=\left\{x=\dfrac{1}{2k+1}|k\in N;0\le k\le1007\right\}\\ C.có:\left(2015-1\right):2+1=1008\left(phần.tử\right)\\ D=\left\{x=\dfrac{1}{2k+1}|k\in N;6\le k\le1010\right\}\\ D.có:\left(2021-13\right):2+1=1005\left(phần.tử\right)\)
TT
2
AH
Akai Haruma
Giáo viên
23 tháng 10
1/
$M=\frac{1}{3.8}+\frac{1}{8.13}+\frac{1}{13.18}+....+\frac{1}{33.38}$
$5M=\frac{8-3}{3.8}+\frac{13-8}{8.13}+\frac{18-13}{13.18}+...+\frac{38-33}{33.38}$
$=\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+....+\frac{1}{33}-\frac{1}{38}$
$=\frac{1}{3}-\frac{1}{38}=\frac{35}{114}$
$\Rightarrow M=\frac{35}{114}:5=\frac{7}{114}$
AH
Akai Haruma
Giáo viên
23 tháng 10
2/
$N=\frac{1}{3.10}+\frac{1}{10.17}+\frac{1}{17.24}+\frac{1}{24.31}+\frac{1}{31.38}$
$7N=\frac{10-3}{3.10}+\frac{17-10}{10.17}+\frac{24-17}{17.24}+\frac{31-24}{24.31}+\frac{38-31}{31.38}$
$=\frac{1}{3}-\frac{1}{10}+\frac{1}{10}-\frac{1}{17}+\frac{1}{17}-\frac{1}{24}+\frac{1}{24}-\frac{1}{31}+\frac{1}{31}-\frac{1}{38}$
$=\frac{1}{3}-\frac{1}{38}=\frac{35}{114}$
$\Rightarrow N=\frac{35}{114}:7=\frac{5}{114}$
VM
1
AH
Akai Haruma
Giáo viên
2 tháng 12 2023
5,1+(x+1):5,1+(x+2):5,1+...+(x+9):5,1=9,5$
$5,1+(x+1+x+2+x+3+...+x+9):5,1=9,5$
$5,1+(9\times x+45):5,1=9,5$
$(9\times x+45):5,1=9,5-5,1=4,4$
$9\times x+45=4,4\times 5,1=22,44$
$9\times x=22,44-45$
Lớp 5 chưa học trừ số bé cho số lớn. Bạn xem lại đề.
VM
1
AH
Akai Haruma
Giáo viên
2 tháng 12 2023
Lời giải:
$5,1+(x+1):5,1+(x+2):5,1+..+(x+9):5,1=9,5$
$5,1+(x+1+x+2+x+3+....+x+9):5,1=9,5$
$5,1+(9\times x+45):5,1=9,5$
$(9\times x+45):5,1=9,5-5,1=4,4$
$9\times x+45=4,4\times 5,1=22,44$
$9\times x=22,44-45$
Lớp 5 chưa học lấy số bé trừ số lớn nên bạn xem lại đề.
TP
2
NX
27 tháng 8 2023
\(D=\left(\dfrac{1}{4}+\dfrac{1}{24}+\dfrac{1}{124}\right):\left(\dfrac{3}{4}+\dfrac{3}{24}+\dfrac{3}{124}\right)+\left(\dfrac{2}{7}+\dfrac{2}{17}+\dfrac{2}{127}\right):\left(\dfrac{3}{7}+\dfrac{3}{17}+\dfrac{3}{127}\right)\)
\(D=\left(\dfrac{1}{4}+\dfrac{1}{24}+\dfrac{1}{124}\right):3\left(\dfrac{1}{4}+\dfrac{1}{24}+\dfrac{1}{124}\right):3\left(\dfrac{1}{7}+\dfrac{1}{27}+\dfrac{1}{127}\right):3\left(\dfrac{1}{7}+\dfrac{1}{27}+\dfrac{1}{127}\right)\)
\(D=\dfrac{1}{3}+\dfrac{2}{3}\)
\(D=1\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
27 tháng 8 2023
D = \(\dfrac{\dfrac{1}{4}+\dfrac{1}{24}+\dfrac{1}{124}}{\dfrac{3}{4}+\dfrac{3}{24}+\dfrac{3}{124}}\) + \(\dfrac{\dfrac{2}{7}+\dfrac{2}{17}+\dfrac{2}{127}}{\dfrac{3}{7}+\dfrac{3}{17}+\dfrac{3}{127}}\)
D = \(\dfrac{\dfrac{1}{4}+\dfrac{1}{24}+\dfrac{1}{124}}{3.\left(\dfrac{1}{4}+\dfrac{1}{24}+\dfrac{1}{124}\right)}\) + \(\dfrac{2.\left(\dfrac{1}{7}+\dfrac{1}{17}+\dfrac{1}{127}\right)}{3.\left(\dfrac{1}{7}+\dfrac{1}{17}+\dfrac{1}{127}\right)}\)
D = \(\dfrac{1}{3}\) + \(\dfrac{2}{3}\)
D = \(\dfrac{3}{3}\)
D = 1
NP
3
TN
17 tháng 10 2016
\(\frac{41}{33}\)\(\frac{1043}{333}\)\(\frac{461}{90}\)\(\frac{73}{3300}\)
A
17 tháng 10 2016
\(1,\left(24\right)\)\(=\)\(1\frac{24}{99}\)\(=\)\(\frac{123}{99}\)
\(3,\left(132\right)\)\(=\)\(3\frac{132}{999}\)\(=\)\(\frac{3129}{999}\)
\(5,1\left(2\right)\)\(=\)\(5\frac{12-1}{90}\)\(=\)\(5\frac{11}{90}\)\(=\)\(\frac{461}{90}\)
\(0,02\left(21\right)\)\(=\)\(\frac{221-2}{990}\)\(=\)\(\frac{219}{990}\)
2 tháng 6 2020
\(A=(1\frac{1}{5}+\frac{17}{24}+\frac{29}{32}+\frac{1}{4}+\frac{7}{24}+\frac{3}{32})\div4\)
\(A=\left\{1\frac{1}{5}+\frac{1}{4}+(\frac{17}{24}+\frac{7}{24})+(\frac{29}{32}+\frac{3}{32})\right\}\div4\)
\(A=\left\{\frac{29}{20}+1+1\right\}\div4\)
\(A=\frac{69}{20}\div4\)
\(A=\frac{69}{80}=0,8625\)
2 tháng 6 2020
a=(6/5+17/24+29/32+1/4+7/24+3/32):4
a=[6/5+(17/24+7/24)+(29/32+3/32+1/4)]:4
a=[6/5+1+1+1/4):4
a=[24/20+20/20+20/20+5/20]:4
a=[44/20+25/20]:4
a=69/20:4
a=69/80
a=
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 - \(\dfrac{97}{7}\)
= - \(\dfrac{2}{5}\) - \(\dfrac{97}{7}\)
= - \(\dfrac{499}{35}\)
1 1/17 . 1 1/24 . (-5,1)=\(-\frac{121}{80}\)