\(\frac{2x-4,36}{0,125}=0,25.42,9-11,7.0,25+0,25.0,8\)

\(\Leftrightarrow\frac{2x-4,36}{0,125}=0,25.\left(42,9-11.7+0,8\right)\)

\(\Leftrightarrow\frac{2x-4,36}{0,125}=0,25.32\)

\(\Leftrightarrow\frac{2x-4,36}{0,125}=8\)

\(\Leftrightarrow2x-4,36=1\)

\(\Leftrightarrow2x=5,36\)

\(\Leftrightarrow x=2,68\)

b) \(N=\frac{1}{1.5}+\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+...+\frac{1}{2005.2010}\)

\(\Leftrightarrow N=\frac{1}{5}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+...+\frac{1}{2005}-\frac{1}{2010}\right)\)

\(\Leftrightarrow N=\frac{1}{5}\left(1-\frac{1}{2010}\right)\)

\(\Leftrightarrow N=\frac{1}{5}.\frac{2009}{2010}=\frac{2009}{10050}\)

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) \(T=\frac{2009\cdot2010+2000}{2011\cdot2010-2020}\)

\(T=\frac{2009\cdot2010+2000}{2009\cdot2010+4020-2020}\)

\(T=\frac{2009\cdot2010+2000}{2009\cdot2010+2000}=1\)

\(N=\frac{1}{1.5}+\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+...+\frac{1}{2005.2010}\)

\(N=\frac{1}{5}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+...+\frac{1}{2005}-\frac{1}{2010}\right)\)

\(N=\frac{1}{5}.\left(1-\frac{1}{2010}\right)\)

\(N=\frac{1}{5}.\frac{2009}{2010}\)

\(N=\frac{2009}{10050}\)

A=1.5.(3.2)+2.10.(6.2)+3.15.(9.2)+4.20.(12.2)+5.25.(15.2)

1.3.5+2.6.10+3.9.15+4.12.20+5.15.25

A=1.5.3+2.10.6+3.15.9+4.20.12+5.25.15(2.2.2.2.2)

1.3.5+2.6.10+3.9.15+4.12.20+5.15.25

A=2.2.2.2.2

A=32

\(\frac{1\cdot3\cdot5\cdot2+2\cdot10\cdot6\cdot2+3\cdot15\cdot9\cdot2+4\cdot20\cdot12\cdot2+5\cdot25\cdot15\cdot2}{1\cdot3\cdot5+2\cdot10\cdot6+3\cdot15\cdot9+4\cdot20\cdot12+5\cdot25\cdot15 }\)

\(2\cdot2\cdot2\cdot2\cdot2=2^5\)

\(=32\)

làm bừa thui,ai trên 11 điểm tích mình mình tích lại

Số số hạng là : 

Có số cặp là :

50 : 2 = 25 ( cặp )

Mỗi cặp có giá trị là :

99 - 97 = 2 

Tổng dãy trên là :

25 x 2 = 50

Đáp số : 50

=\(\frac{30\left(1+8+27+64+125\right)}{5\left(3+24+81+64+375\right)}\)

= \(\frac{30.225}{5.574}\)

=\(\frac{6750}{2870}\)

=\(\frac{675}{287}\)

K mình với!!!!

a) \(\frac{1}{2}-\frac{1}{3.7}-\frac{1}{7.11}-\frac{1}{11.15}-\frac{1}{15.19}-\frac{1}{19.23}-\frac{1}{23.27}\)

\(=\frac{1}{2}-\left(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+\frac{1}{15.19}+\frac{1}{19.23}+\frac{1}{23.27}\right)\)

\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+\frac{1}{19}-\frac{1}{19}+\frac{1}{23}-\frac{1}{23}+\frac{1}{27}\right)\)

\(=\frac{1}{2}-\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{27}\right)\)

\(=\frac{1}{2}-\frac{1}{4}.\frac{8}{27}\)

\(=\frac{1}{2}-\frac{2}{27}\)

\(=\frac{23}{54}\)

b) \(1-\frac{1}{5.10}-\frac{1}{10.15}-\frac{1}{15.20}-...-\frac{1}{95.100}\)

\(=1-\left(\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+...+\frac{1}{95.100}\right)\)

\(=1-\frac{1}{5}.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+\frac{1}{20}-...-\frac{1}{95}-\frac{1}{100}\right)\)

\(=1-\frac{1}{5}.\left(\frac{1}{5}-\frac{1}{100}\right)\)

\(=1-\frac{1}{5}.\frac{19}{100}\)

\(=1-\frac{19}{500}\)

\(=\frac{481}{500}\)

\(=\frac{219}{520}=\frac{155052}{368160}\)

\(=\frac{303}{708}=\frac{157560}{368160}\)

\(\frac{155052}{368160}< \frac{157560}{368160}\)

VẬY \(\frac{303}{708}\)LỚN HƠN

\(\frac{1.5.6+2.10.12+4.20.24+9.45.54}{1.3.5+2.6.10+4.12.20+9.27.45}\)

\(=\frac{1.5.3.2+2.10.6.2+4.20.12.2+9.45.27.2}{1.3.5+2.6.10+4.12.20+9.27.45}\)

\(=\frac{2\left(1.3.5+2.6.10+4.12.20+9.27.45\right)}{1.3.5+2.6.10+4.12.20+9.27.45}\)

\(=2\)