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TT
31 tháng 7 2020
\(\frac{2x-4,36}{0,125}=0,25.42,9-11,7.0,25+0,25.0,8\)
\(\Leftrightarrow\frac{2x-4,36}{0,125}=0,25.\left(42,9-11.7+0,8\right)\)
\(\Leftrightarrow\frac{2x-4,36}{0,125}=0,25.32\)
\(\Leftrightarrow\frac{2x-4,36}{0,125}=8\)
\(\Leftrightarrow2x-4,36=1\)
\(\Leftrightarrow2x=5,36\)
\(\Leftrightarrow x=2,68\)
b) \(N=\frac{1}{1.5}+\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+...+\frac{1}{2005.2010}\)
\(\Leftrightarrow N=\frac{1}{5}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+...+\frac{1}{2005}-\frac{1}{2010}\right)\)
\(\Leftrightarrow N=\frac{1}{5}\left(1-\frac{1}{2010}\right)\)
\(\Leftrightarrow N=\frac{1}{5}.\frac{2009}{2010}=\frac{2009}{10050}\)
31 tháng 7 2020
Bài 1:
a)\(\frac{2\cdot x-4,36}{0,125}=0,25\cdot42,9-11,7\cdot0,25+0,25\cdot0,8\)
\(\frac{2\cdot x-4,36}{0,125}=0,25\cdot\left(42,9-11,7+0,8\right)\)
\(\frac{2\cdot x-4,36}{0,125}=0,25\cdot32\)
\(\frac{2\cdot x-4,36}{0,125}=8\)
\(2\cdot x-4,36=8\cdot0,125\)
\(2\cdot x-4,36=1\)
\(2\cdot x=1+4,36\)
\(2\cdot x=5,36\)
\(x=\frac{5,36}{2}=2,68\)
b) \(N=\frac{1}{1\cdot5}+\frac{1}{5\cdot10}+\frac{1}{10\cdot15}+\frac{1}{15\cdot20}+...+\frac{1}{2005\cdot2010}\)
\(4N=\frac{4}{1\cdot5}+\frac{4}{5\cdot10}+\frac{4}{10\cdot15}+\frac{4}{15\cdot20}+...+\frac{4}{2005\cdot2010}\)
\(4N=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+...+\frac{1}{2005}-\frac{1}{2010}\)
\(4N=1-\frac{1}{2010}=\frac{2009}{2010}\)
\(N=\frac{2009}{2010}\div4=\frac{2009}{8040}\)
Bài 2:
a) ( x + 5,2 ) : 3,2 = 4,7 ( dư 0,5 )
\(x+5,2=4,7\cdot3,2+0,5\)
\(x+5,2=15,54\)
\(x=15,54-5,2=10,34\)
b)\(A=\frac{4047991-2010\cdot2009}{4050000-2011\cdot2009}\)
\(A=\frac{4047991-2010\cdot2009}{4050000-2009-2010\cdot2009}\)
\(A=\frac{4047991-2010\cdot2009}{4047991-2010\cdot2009}=1\)
Bài 3:
a) \(104,5\cdot x-14,1\cdot x+9,6\cdot x=25\)
\(x\cdot\left(104,5-14,1+9,6\right)=25\)
\(x\cdot100=25\)
\(x=\frac{25}{100}=\frac{1}{4}=0,25\)
b) \(T=\frac{2009\cdot2010+2000}{2011\cdot2010-2020}\)
\(T=\frac{2009\cdot2010+2000}{2009\cdot2010+4020-2020}\)
\(T=\frac{2009\cdot2010+2000}{2009\cdot2010+2000}=1\)
M
2
1 tháng 8 2021
Bài 1 :
a) 8,6 * x + x * 1,4 = 10
=> x * ( 8,6 + 1,4 ) = 10
=> x * 10 = 10
=> x = 1
1 tháng 8 2021
b) 2009 x 2009 + 2010
= 2009 x 2009 x ( 2009 + 1 )
= 2009 x 2009 x 2009 + 2009 x 2009
= 20093 + 20092
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, \(\frac{1}{3}\)+ \(\frac{1}{15}\)+ \(\frac{1}{35}\)+ \(\frac{1}{63}\)+ \(\frac{1}{99}\)+ \(\frac{1}{143}\)+ \(\frac{1}{195}\)
= \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{5}\)+ \(\frac{1}{5}\)- \(\frac{1}{7}\)+\(\frac{1}{7}\)- \(\frac{1}{9}\)+...........+\(\frac{1}{13}\)- \(\frac{1}{15}\)
= \(\frac{1}{3}\)- \(\frac{1}{15}\)
= \(\frac{4}{15}\)
29 tháng 7 2015
=> \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{2010}{2011}\)
=> \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2010}{2011}\)
=>\(1-\frac{1}{x+1}=\frac{2010}{2011}\)
=> \(\frac{1}{x+1}=\frac{2011}{2011}-\frac{2010}{2011}=\frac{1}{2011}\)
=> x + 1 = 2011
=> x = 2010
8 tháng 6 2017
\(\frac{20102010}{20112011}=\frac{2010}{2011}\)
\(\frac{20122012}{20112011}=\frac{2012}{2011}\)
\(\left(\frac{2010}{2011}+\frac{2012}{2011}\right).x-1=2011\)
\(\frac{4022}{2011}.x-1=2011\)
\(2.x-1=2011\)
\(2.x=2011+1\)
\(2.x=2012\)
\(x=2012:2\)
\(x=1006\)
DQ
3
3 tháng 9 2016
\(\left(\frac{2010:x-6}{2011}+\frac{3}{4}\right)x\frac{11}{3}=1\)
\(\frac{2010:x-6}{2011}+\frac{3}{4}=\frac{3}{11}\)
\(\frac{2010:x-6}{2011}=-\frac{21}{44}\)
\(\Rightarrow44\left(2010:x-6\right)=\left(-21\right).2011\)
\(\Rightarrow88440:x-264=-42231\)
\(\Rightarrow88440:x=-41967\)
\(\Rightarrow x=\frac{88440}{-41967}\approx2,108\)
3 tháng 9 2016
\(\left(\frac{2010x-6}{2011}+\frac{3}{4}\right)\times\frac{11}{13}=1\)
\(\frac{2010x-6}{2011}+\frac{3}{4}=\frac{11}{13}\)
\(\frac{2010x-6}{2011}=\frac{11}{13}-\frac{3}{4}\)
\(\frac{2010x-6}{2011}=\frac{5}{52}\)
\(\left(2010x-6\right)\div2011=\frac{5}{52}\)
\(2010x-6=\frac{5}{52}\times2011\)
\(2010x-6=\frac{10055}{52}\)
\(2010x=\frac{10055}{52}+6\)
\(2010x=\frac{10367}{52}\)
\(x=\frac{10367}{52}:2010\)
\(x=\frac{10367}{104520}\)
5 x y -1952 =553
5 x y = 553 + 1952
5 x y = 2505
y = 2505 : 5
y = 501
y x 2011 - y = 2011 x 2009 + 2011 x 1
y x 2011 - y x 1 =2011 x 2009 + 2011x1
y x ( 2011 - 1) = 2011 x ( 2009 +1)
y x 2010 = 2011 x 2010
nhìn vào phép tính trên ta thấy cả 2 vế đều có 1 thừa số chung đó là 2010
mà 2 vế bằng nhau
nên y = 2011
bn nào đi qua thấy đúng cho mk 1 k nha. yêu mọi người moa
1
a)
5 x X - 1952 = 2500 - 1947
5 x X - 1952 = 553
5 x X = 553 + 1952
5 x X = 2505
X = 2505 : 5
X = 501
b)
X x 2011 - X = 2011 x 2009 + 2011
X x 2011 - X x 1 = 2011 x 2009 + 2011 x 1
X x ( 2011 + 1 ) = 2011 x ( 2009 + 1 )
X x 2012 = 2011 x 2010
X x 2012 = 4042110
X = 4042110 : 2012
X = \(_{\frac{2021055_{ }}{1006}}\)