tinh nhanh
A=\(\frac{4047991-2010x2009}{4050000-2011x2009}\)
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VT
1
DD
Đoàn Đức Hà
Giáo viên
18 tháng 5 2021
\(A=\frac{4047991-2010\times2009}{4050000-2011\times2009}=\frac{4047991+2009-2009-2010\times2009}{4050000-2011\times2009}\)
\(=\frac{4050000-2011\times2009}{4050000-2011\times2009}=1\)
NP
0
24 tháng 7 2015
4047991- 2010 . 2009 / 4050000 - 2011 . 2009
=9901/9901
=1
HN
4
S
27 tháng 5 2019
\(A=\frac{4047991-2010.2009}{4050000-2011.2009}\)
\(\Rightarrow A=\frac{404791-2010.2009}{4047911+2009-2011.2009}\)
\(\Rightarrow A=\frac{4047911-2010.2009}{4047911-2010.2009}\)
\(\Rightarrow A=1\)
Vậy A = 1
~Study well~
#๖ۣۜNamiko#
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) \(T=\frac{2009\cdot2010+2000}{2011\cdot2010-2020}\)
\(T=\frac{2009\cdot2010+2000}{2009\cdot2010+4020-2020}\)
\(T=\frac{2009\cdot2010+2000}{2009\cdot2010+2000}=1\)
7 tháng 4 2021
a) Ta có: 2,735+1,248+3,265+4,752
=(2,735+3,265)+(1,248+4,752)
=6+6=12
b) Ta có: (10,38+12,56+14,68)-(0,38+4,68+2,58)
=(10,38-0,38)+(12,56-2,58)+(14,68-4,68)
=10+10+9,98
=29,98
\(A=\frac{4047991-2010.2009}{4050000-2011.2009}=\frac{4047991-2010.2009}{4050000-\left(2010+1\right).2009}=\frac{4047991-2010.2009}{4050000-\left(2010.2009+1.2009\right)}\)
\(=\frac{4047991-1}{4050000-\left(1+2009\right)}=\frac{4047990}{4050000-2010}=\frac{4047990}{4047990}=1\)