x/3+x/15+x/35+................+x/99+x/143=6/13
có giải cho tao không
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NV
2
VA
28 tháng 7 2016
Ta có :
Đặt \(A=\frac{31}{3}+\frac{31}{15}+...+\frac{31}{143}\)
\(A=\frac{31}{1.3}+\frac{31}{3.5}+...+\frac{31}{11.13}\)
\(\frac{2}{31}A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{11.13}\)
\(\frac{2}{31}A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{13}\)
\(\frac{2}{31}A=1-\frac{1}{13}\)
\(A=\frac{12}{13}:\frac{2}{31}\)
\(A=\frac{186}{13}\)
\(\Rightarrow x-\frac{186}{13}=\frac{9}{13}\)
\(x=\frac{9}{13}+\frac{186}{13}\)
\(x=\frac{195}{13}=15\)
Ủng hộ mk nha !!! ^_^
18 tháng 8 2021
\(\text{Đặt }A=\frac{31}{3}+\frac{31}{15}+...+\frac{31}{143}\)
\(\Rightarrow A=\frac{31}{1.3}+\frac{31}{3.5}+...+\frac{31}{11.13}\)
\(\Rightarrow\frac{2}{31}A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{11.13}\)
\(\Rightarrow\frac{2}{31}A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{13}\)
\(\Rightarrow\frac{2}{31}A=1-\frac{1}{13}\Rightarrow A=\frac{12}{13}:\frac{2}{31}=\frac{186}{13}\)
\(\Rightarrow x-\frac{186}{13}=\frac{9}{13}\)
\(\Rightarrow x=\frac{9}{13}+\frac{186}{13}\)
\(\Rightarrow x=\frac{195}{13}=15\)
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
PM
1
6 tháng 2 2016
a) (x-2)(x+3) <0 => x-2 và x+3 phải trái dấu
=> x-2<0 và x+3>0
hoặc x-2>0 và x+3<0
=> x<2 và x>-3 => -3<x<2
hoặc x>2 và x<-3 ( vô lý ) ( loại )
=> x \(\in\) { -2;-1;0;1 }
Đúng 100%, tích nha, please!!
TT
18 tháng 7 2016
Ta có: x-(\(\frac{31}{5}+\frac{31}{3.5}+\frac{31}{5.7}+\frac{31}{7.9}+\frac{31}{9.11}\)\(+\frac{31}{11.13}\))=\(\frac{9}{13}\)
x-\(\frac{31}{5}\)-\(\frac{31}{2}\)x(\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\))=\(\frac{9}{13}\)
x-\(\frac{31}{5}-\frac{31}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{11}-\frac{1}{13}\right)\)=\(\frac{9}{13}\)
x-\(\frac{31}{5}\)\(-\frac{31}{2}\left(\frac{1}{3}-\frac{1}{13}\right)=\frac{9}{13}\)
x-\(\frac{31}{5}-\frac{31}{2}.\frac{10}{39}\)\(=\frac{9}{13}\)
x-\(\frac{31}{5}-\frac{155}{39}=\frac{9}{13}\)
x-\(\frac{434}{195}\)=\(\frac{9}{13}\)
x =\(\frac{9}{13}+\frac{434}{195}=\frac{569}{195}\)
nhé
TT
0
25 tháng 6 2018
\(\frac{31}{3}+\frac{31}{15}+\frac{31}{35}+\frac{31}{63}+\frac{31}{99}+\frac{31}{143}=\frac{31}{1.3}+\frac{31}{3.5}+\frac{31}{5.7}+\frac{31}{7.9}+\frac{31}{9.11}+\frac{31}{11.13}\\ \)
\(=\frac{31}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(=\frac{31}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{31}{2}.\left(1-\frac{1}{13}\right)=\frac{31}{2}.\frac{12}{13}=\frac{31.6}{13}=\frac{186}{13}\)
\(\Rightarrow x-\frac{186}{13}=\frac{9}{13}\Leftrightarrow x=\frac{195}{13}=15\)
NV
25 tháng 6 2018
\(x-\left(\frac{31}{3}+\frac{31}{15}+\frac{31}{35}+\frac{31}{63}+\frac{31}{99}+\frac{31}{143}\right)=\)\(\frac{9}{13}\)(1)
Đặt \(A=\frac{31}{3}+\frac{31}{15}+\frac{31}{35}+\frac{31}{63}+\frac{31}{99}+\frac{31}{143}\)
\(A=31\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\right)\)
\(\Rightarrow2A=31\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\right)\)
\(2A=31\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(2A=31\left(2-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(2A=31\left(2-\frac{1}{13}\right)\)
\(2A=31.\frac{25}{13}\)
\(2A=\frac{775}{13}\)
\(\Rightarrow A=\frac{775}{13}:2\)
\(A=\frac{775}{26}\)
Thay vào (1) ta có:
\(x-\frac{775}{26}=\frac{9}{13}\)
\(\Leftrightarrow x=\frac{9}{13}+\frac{775}{26}\)
\(\Leftrightarrow x=\frac{61}{2}\)
\(\frac{x}{1.3}+\frac{x}{3.5}+\frac{x}{5.7}+.....+\frac{x}{9.11}+\frac{x}{11.13}=\frac{6}{13}\)
=>\(\frac{x}{1}-\frac{x}{3}+\frac{x}{3}-\frac{x}{5}+\frac{x}{5}-\frac{x}{7}+......+\frac{x}{9}-\frac{x}{11}+\frac{x}{11}-\frac{x}{13}=\frac{12}{13}\)
\(\frac{x}{1}-\frac{x}{13}=\frac{12}{13}\)
x.12/13=12/13 => x=1