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26 tháng 4 2020
(X - 1005) x (1 + 3 + 5+ 7 + 9 + 11) = 1989 x 1990 x (70 - 35 x 2)
(X - 1005) x (11 + 1)[(11 - 1) : 2 + 1): 2] = 1989 x 1990 x 0
(X - 1005) x 36 = 0
X - 1005 = 0 : 36
X - 1005 = 0
X = 1005
26 tháng 4 2020
( x - 1005 ) . ( 1 + 3 + 5 + 7 + 9 + 11 ) = 1989 . 1990 . ( 70 - 35 . 2 )
( x - 1005 ) . 36 = 1989 . 1990 . ( 70 - 70 )
( x - 1005 ) . 36 = 1989 . 1990 . 0
( x - 1005 ) . 36 = 0
x - 1005 = 0
x = 1005
XO
1 tháng 7 2019
a) \(\frac{1}{3}+\frac{5}{6}:\left(x-2\frac{1}{5}\right)=\frac{3}{4}\)
=> \(\frac{1}{3}+\frac{5}{6}:\left(x-\frac{11}{5}\right)=\frac{3}{4}\)
=> \(\frac{5}{6}:\left(x-\frac{11}{5}\right)=\frac{3}{4}-\frac{1}{3}\)
=> \(\frac{5}{6}:\left(x-\frac{11}{5}\right)=\frac{5}{12}\)
=> \(x-\frac{11}{5}=\frac{5}{6}:\frac{5}{12}\)
=> \(x-\frac{11}{5}=2\)
=> \(x=2+\frac{11}{5}\)
=> \(x=\frac{21}{5}\)
NA
Tìm x :
(x - 21*13) : 11 = 30
(627 - 56,5) : (x: 2) = 163
2/3 + 1/3 : x = 1
165 - (35 : x + 3) * 15 = 15
5
28 tháng 4 2018
a, ( x - 21 . 13 ) : 11 = 30
x - 273 = 30 . 11
x - 273 = 330
x = 330 + 273
x = 603
b, ( 627 - 56,5 ) : ( x : 2 ) = 163
570,5 : ( x : 2 ) = 163
x : 2 = 570,5 : 163
x : 2 = 3,5
x = 3,5 . 2
x = 7
c, 2/3 + 1/3 : x = 1
1/3 : x = 1 - 2/3
1/3 : x = 1/3
x = 1/3 : 1/3
x = 1
d, 165 - ( 35 : x + 3 ) . 15 = 15
165 - ( 35 : x + 3 ) = 15 : 15
165 - ( 35 : x + 3 ) = 1
35 : x + 3 = 165 - 1
35 : x + 3 = 164
35 : x = 164 - 3
35 : x = 161
x = 35 : 161
x = 5/23
Mk tính nhẩm nên chắc có chỗ sai nên bn cứ kiểm tra lại nha
N
25 tháng 7 2018
a) ( 1 + 3 + 5 + 7 + ....... + 2007 + 2009 + 2011 ) x ( 125125 x 127 - 127127 x 125 )
vì ( 125125 x 127 - 127127 x 125 ) =[125125 x (125+2)] - 127127 x 125 ) =>125125 x (125+2)=125.125125+125125.2=125125.125+250250=125125.125+125.2002=125.(125125+2002)=125.127127
=> ( 125125 x 127 - 127127 x 125 )=127127.125-127127.125=0
=> (1 + 3 + 5 + 7 + ....... + 2007 + 2009 + 2011 ) x ( 125125 x 127 - 127127 x 125 ) =0
25 tháng 7 2018
a) ( 1 + 3 + 5 + 7 + ....... + 2007 + 2009 + 2011 ) x ( 125125 x 127 - 127127 x 125 )
= ( 1 + 3 + 5 + 7 + ....... + 2007 + 2009 + 2011 ) x 0
= 0
b, \(\frac{1}{3}\)+ \(\frac{1}{15}\)+ \(\frac{1}{35}\)+ \(\frac{1}{63}\)+ \(\frac{1}{99}\)+ \(\frac{1}{143}\)+ \(\frac{1}{195}\)
= \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{5}\)+ \(\frac{1}{5}\)- \(\frac{1}{7}\)+\(\frac{1}{7}\)- \(\frac{1}{9}\)+...........+\(\frac{1}{13}\)- \(\frac{1}{15}\)
= \(\frac{1}{3}\)- \(\frac{1}{15}\)
= \(\frac{4}{15}\)
23 tháng 2 2016
1/3+1/15+1/35+1/63+1/99+1/143+1/195
=1/1*3+1/3*5+1/5*7+1/7*9+1/9*11+1/11*13+1/13*15
suy ra 2(1/1*3+1/3*5+1/5*7+1/7*9+1/9*11+1/11*13+1/13*15)
=2/1*3+2/3*5+2/5*7+2/7*9+2/9*11+2/11*13+2/13*15
=1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13+1/13-1/15
=1-1/15
=14/15
a=14/15 chia 2=7/15
TC
0
24 tháng 6 2018
a) x + 3.x + 5.x + ... + 2009. x = 2010.1005
\(x\times\left(1+3+5+...+2009\right)=2010\times1005\)
\(x\times\left[\left(1+2009\right)\times1005:2\right]=2010\times1005\)
\(x\times2010\times1005\times\frac{1}{2}=2010\times1005\)
\(\Rightarrow x\times\frac{1}{2}=2010\times1005:\left(2010\times1005\right)\)
\(x\times\frac{1}{2}=1\)
x = 2
b) x + (x+1) + (x+2) +...+ (x+30) = 620
x. 31 + ( 1+2+...+30) = 620
x.31 + [ ( 30+1).30:2) = 620
x.31 + 465 = 620
x.31 = 620 - 465
x.31 = 155
x = 155 : 31
x = 5
24 tháng 6 2018
a) x+3.x+5.x+.....+2009.x = 2010.1005
=> x.(1+3+5+....+2009) = 2010.1005
=> x.1010025 = 2020050
=> x = 2
Vậy x = 2
b) x+(x+1)+(x+2)+....+(x+30) = 620
=> (x+x+x+...+x)+(1+2+3+...+30) = 620
=> 31x + 465 = 620
=> 31x = 155
=> x = 5
\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+....+\frac{1}{x\times\left(x+2\right)}=\frac{1005}{2011}\)
\(\Leftrightarrow\frac{1}{2}\times\left(\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+.....+\frac{1}{x\times\left(x+2\right)}\right)=\frac{1005}{2011}\)
\(\Leftrightarrow\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{1005}{2011}\)
\(\Leftrightarrow\frac{1}{2}\times\left(1-\frac{1}{x+2}\right)=\frac{1005}{2011}\)
\(\Leftrightarrow1-\frac{1}{x+2}=\frac{1005}{2011}:\frac{1}{2}=\frac{2010}{2011}\)
\(\Leftrightarrow\frac{1}{x+2}=1-\frac{2010}{2011}=\frac{1}{2011}\)
\(\Leftrightarrow x+2=2011\)
\(\Leftrightarrow x=2009\)
Vậy x = 2009
\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{x\left(x+2\right)}=\frac{1005}{2011}\)
\(\Rightarrow\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{1005}{2011}\)
\(\Rightarrow\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{1005}{2011}\)
\(\Rightarrow\frac{1}{2}\left(1-\frac{1}{x+2}\right)=\frac{1005}{2011}\)
\(\Rightarrow\frac{x+1}{x+2}=\frac{2010}{2011}\)
\(\Rightarrow x+1=2010\Leftrightarrow x=2009\)