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MH
4
T
9 tháng 9 2018
\(\frac{3}{4}+\frac{3}{28}+\frac{3}{70}+\frac{3}{130}+....+\frac{3}{304}.\)
\(=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+.....+\frac{3}{16.19}\)
\(=\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+\frac{13-10}{10.13}+.....+\frac{19-16}{16.19}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+....+\frac{1}{16}-\frac{1}{19}\)
\(=1-\frac{1}{19}=\frac{18}{19}\)
9 tháng 9 2018
3/4+3/28+3/70+3/130+...+3/304
= 3 /1.4 + 3/4.7 + 3/7.10 + 3/10.13 +....+ 3 /16.19
= 1 -1/4 + 1/4 - 1/7 + 1/7 - 1/10 + 1/10 - 1/13 + ...+1/16 - 1/19
= 1 - 1/19
=18/19
DT
Dang Tung
CTVHS
23 tháng 8 2023
\(\dfrac{3}{4}+\dfrac{3}{28}+\dfrac{3}{70}+\dfrac{3}{130}+\dfrac{3}{208}+\dfrac{3}{304}\\ =\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+\dfrac{3}{13.16}+\dfrac{3}{16.19}\\ =1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{19}\\ =1-\dfrac{1}{19}=\dfrac{18}{19}\)
TN
23 tháng 8 2023
\(\dfrac{3}{4}+\dfrac{3}{28}+\dfrac{3}{70}+\dfrac{3}{130}+\dfrac{3}{208}+\dfrac{3}{304}\)
\(=\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+\dfrac{3}{7\times10}+\dfrac{3}{10\times13}+\dfrac{3}{13\times16}+\dfrac{3}{16\times19}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{19}\)
\(=1-\dfrac{1}{19}\)
=\(\dfrac{18}{19}\)
NT
4
8 tháng 9 2017
A=\(\frac{3}{4}+\frac{3}{28}+\frac{3}{70}+\frac{3}{130}+\frac{3}{208}+\frac{3}{304}\)
A= \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}+\frac{3}{16.19}\)
A= \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{19}\)
A= 1 - \(\frac{1}{19}\)
A= \(\frac{18}{19}\)
8 tháng 9 2017
X=3/4+3/28+...+3/304
X=3/1x4+3/4x7+....+3/13x16
X=1-1/4+1/4-1/7+....+1/13-1/16
X=1-1/16
X=15/16
k cho tớ nha 
Nguyễn Thu Trang
T
2 tháng 7 2019
#)Giải :
\(200-18:\left(372:3x-1\right)-28=166\)
\(\Leftrightarrow200-18:\left(372:3x-1\right)=194\)
\(\Leftrightarrow18:\left(372:3x-1\right)=6\)
\(\Leftrightarrow372:3x-1=3\)
\(\Leftrightarrow3x-1=124\)
\(\Leftrightarrow3x=125\)
\(\Leftrightarrow x=\frac{125}{3}\)
XO
2 tháng 7 2019
200 - 18 : (372 : 3 . x - 1) - 28 = 166
=> 200 - 18 : (372 : 3.x - 1) = 166 + 28
=> 200 - 18 : (372 : 3.x) - 1) = 194
=> 18 : (372 : 3.x - 1) = 200 - 194
=> 18 : (372 : 3.x - 1) = 6
=> 372 : 3.x - 1 = 18 : 6
=> 372 : 3.x - 1 = 3
=> 372 : 3.x = 3 + 1
=> 372 : 3.x = 4
=> 3.x = 372 : 4
=> 3.x = 93
=> x = 93 : 3
=> x = 31
LH
0
H9
HT.Phong (9A5)
CTVHS
24 tháng 9 2023
a) \(\dfrac{6}{13}:\left(\dfrac{1}{2}-x\right)=\dfrac{15}{39}\)
\(\dfrac{1}{2}-x=\dfrac{6}{13}:\dfrac{15}{39}\)
\(\dfrac{1}{2}-x=\dfrac{6}{5}\)
\(x=\dfrac{1}{2}-\dfrac{6}{5}\)
\(x=-\dfrac{7}{10}\)
b) \(3\times\left(\dfrac{x}{4}+\dfrac{x}{28}+\dfrac{x}{70}+\dfrac{x}{130}\right)=\dfrac{60}{13}\)
\(3\times x\times\left(\dfrac{1}{4}+\dfrac{1}{28}+\dfrac{1}{70}+\dfrac{1}{130}\right)=\dfrac{60}{13}\)
\(x\times\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+\dfrac{3}{7\times10}+\dfrac{3}{7\times13}\right)=\dfrac{60}{13}\)
\(x\times\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}\right)=\dfrac{60}{13}\)
\(x\times\left(1-\dfrac{1}{13}\right)=\dfrac{60}{13}\)
\(x\times\dfrac{12}{13}=\dfrac{60}{13}\)
\(x=\dfrac{60}{13}:\dfrac{12}{13}\)
\(x=5\)
LN
1
17 tháng 5 2020
\(\frac{3}{4}+\frac{3}{28}+\frac{3}{70}+\frac{3}{130}+....+\frac{3}{418}+\frac{3}{550}\)
\(\Leftrightarrow\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{19.22}+\frac{3}{22.25}\)
\(\Leftrightarrow\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{19}-\frac{1}{22}+\frac{1}{22}-\frac{1}{25}\)
\(\Leftrightarrow\frac{1}{1}-\frac{1}{25}=\frac{24}{25}\)
Nhớ k cho m nhé!
QT
2
10 tháng 3 2020
\(\frac{3}{4}+\frac{3}{28}+\frac{3}{70}+...+\frac{3}{10300}\)
\(=\frac{3}{1\times4}+\frac{3}{4\times7}+\frac{3}{7\times10}+...+\frac{3}{100\times103}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\)
\(=1-\frac{1}{103}=\frac{102}{103}\)
HC
30 tháng 6 2019
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XO
30 tháng 6 2019
a) Ta có B = \(\left(\frac{2}{15}+\frac{3}{40}+\frac{4}{96}+\frac{5}{204}+\frac{6}{391}\right).x.\left(x-1\right)=\frac{20}{69}\)
=> \(\left(\frac{2}{3.5}+\frac{3}{5.8}+\frac{4}{8.12}+\frac{5}{12.17}+\frac{6}{17.23}\right).x.\left(x-1\right)=\frac{20}{69}\)
=> \(\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{23}\right).x.\left(x-1\right)=\frac{20}{69}\)
=> \(\left(\frac{1}{3}-\frac{1}{23}\right).x.\left(x-1\right)=\frac{20}{69}\)
=> \(\frac{20}{69}.x.\left(x-1\right)=\frac{20}{69}\)
=> \(x.\left(x-1\right)=\frac{20}{69}:\frac{20}{69}\)
=> \(x.\left(x-1\right)=1\)
=> \(x\in\varnothing\)
a) \(\left(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+....+\frac{1}{8554}\right).x=\frac{31}{94}\)
=> \(\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{91.94}\right).x=\frac{31}{94}\)
=> \(\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{91.94}\right)=\frac{31}{94}\)
=> \(\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{91}-\frac{1}{94}\right).x=\frac{31}{94}\)
=> \(\frac{1}{3}.\left(1-\frac{1}{94}\right).x=\frac{31}{94}\)
=> \(\frac{1}{3}.\frac{93}{94}.x=\frac{31}{94}\)
=> \(\frac{31}{94}.x=\frac{31}{94}\)
=> \(x=\frac{31}{94}:\frac{31}{94}\)
=> \(x=1\)