\(A=8400\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}\right)\)

\(=\frac{8400}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+\frac{4}{17.21}+\frac{4}{21.25}\right)\)

\(=2100\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}+\frac{1}{21}-\frac{1}{25}\right)\)

\(=2100\left(1-\frac{1}{25}\right)\)

\(=2100\cdot\frac{24}{25}\)

\(=2016\)

\(A=8400.\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}\right)\)

\(A=8400.\left(\frac{1.4}{1.5.4}+\frac{1.4}{5.9.4}+\frac{1.4}{9.13.4}+\frac{1.4}{13.17.4}+\frac{1.4}{17.21.4}+\frac{1.4}{21.25.4}\right)\)

\(A=8400.\frac{1}{4}.\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}\right)\)

\(A=8400.\frac{1}{4}.\left(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}+\frac{1}{21}-\frac{1}{25}\right)\)

\(A=8400.\frac{1}{4}.\left(\frac{1}{1}-\frac{1}{25}\right)\)

\(A=8400.\frac{1}{4}.\frac{24}{25}\)

\(A=2016\)

\(A=8400.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}+\frac{1}{21}-\frac{1}{25}\right)\) 

\(A=8400.\left(1-\frac{1}{25}\right)\)

\(A=8400.\frac{24}{25}=8064\)

\(->A=8064\)

a) \(=\frac{9}{1.4}+\frac{9}{4.7}+\frac{9}{7.10}+...+\frac{9}{61.64}\)

\(=3\left(\frac{1}{1}-\frac{1}{64}\right)\)

\(=\frac{189}{64}\)

b) \(=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{21}-\frac{1}{25}\)

\(=\frac{1}{1}-\frac{1}{25}\)

\(=\frac{24}{25}\)

c) Chưa học tới

b)1/1.5+1/5.9+1/9.13+...+1/21.25

=1/4.(4/1.5+4/5.9+4/9.13+4/21.25)

=1/4.(4-4/5+4/5-4/9+4/9-4/13+...+4/21-4/25)

=1/4.(4-4/25)

=1/4.(100/25-4/25)

=1/4.96/25

=24/25

\(P=\frac{3}{1.5}+\frac{3}{5.9}+\frac{3}{9.13}+...+\frac{3}{197.201}\)
\(P=\frac{3}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{197.201}\right)\)
\(P=\frac{3}{4}.\left(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}+\frac{1}{13}+...+\frac{1}{197}-\frac{1}{201}\right)\)
\(P=\frac{3}{4}.\left(\frac{1}{1}-\frac{1}{201}\right)\)
\(P=\frac{3}{4}.\left(\frac{201}{201}-\frac{1}{201}\right)\)
\(P=\frac{3}{4}.\frac{200}{201}\)
\(P=\frac{50}{67}\)
 Vậy \(P=\frac{50}{67}\)

\(P=\frac{3}{1\cdot5}+\frac{3}{5\cdot9}+...+\frac{3}{197\cdot201}\)

\(=3\cdot\left(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+...+\frac{1}{197\cdot201}\right)\)

\(=\frac{3}{4}\cdot\left(\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+...+\frac{4}{197\cdot201}\right)\)

\(=\frac{3}{4}\cdot\left(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{197}-\frac{1}{201}\right)\)

\(=\frac{3}{4}\cdot\left(\frac{1}{1}-\frac{1}{201}\right)\)

\(=\frac{3}{4}\cdot\left(\frac{201-1}{201}\right)\)

\(=\frac{3}{4}\cdot\frac{200}{201}\)

\(\Rightarrow B=\frac{50}{67}\)

\(\frac{4}{3}\) x P = \(\frac{4}{1x5}\)+ \(\frac{4}{5x9}\)+ ...... + \(\frac{4}{197x201}\)

\(\frac{4}{3}\)x P = \(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+.....+\frac{1}{197}-\frac{1}{201}\)

\(\frac{4}{3}\)x P = 1 - \(\frac{1}{201}\)

\(\frac{4}{3}\)x P = \(\frac{200}{201}\)

P = \(\frac{200}{201}\) : \(\frac{4}{3}\)= \(\frac{50}{67}\)

3/1*5+3/5*9+3/9*13+.....+3/3993*3997+3/3997*4001

=1/3(1-1/5+1/5-1/9+1/9-1/13+....+1/3993-1/3997+1/3997-1/4001)

=1/3(1-1/4001)

=4000/12003

k nha

= 3/4(1-1/5+1/5-1/9+1/9-1/13+...+1/3993-1/3997+1/3997-1/4001)

=3/4(1-1/4001)

=3000/4001

\(A=3\times\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{101}-\frac{1}{105}\right)\)

\(A=3\times\left(1-\frac{1}{105}\right)\)

\(A=3\times\frac{104}{105}\)

\(A=\frac{104}{35}\)

- A ở trên giữa các phân số là dấu " + " nha mấy bạn !

Giải:

Ta có công thức sau:

\(\frac{k}{a.b}=\frac{1}{a}-\frac{1}{b}\) với b - a = k hoặc a - b = k

Lắp vào biểu thức A, ta có:

\(A=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.14}+...+\frac{4}{2005.2009}\\ =\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{2001}-\frac{1}{2005}+\frac{1}{2005}-\frac{1}{2009}\)

\(=1+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{9}-\frac{1}{9}\right)+...+\left(\frac{1}{2005}-\frac{1}{2005}\right)-\frac{1}{2009}\\ =1-\frac{1}{2009}\\ =\frac{2009-1}{2009}\\ =\frac{2008}{2009}\)

Vậy \(A=\frac{2008}{2009}\)

Chúc bạn học tốt!

Thanks bạn nha!

4/1.5+4/5.9+4/9.13+…4/2005.2009

=1/1-1/5+1/5-1/9+1/9-1/13+…+1/2005-1/2009

=1-1/2009=2008/2009

\(x+\frac{3}{5.9}+\frac{3}{9.13}+\frac{3}{13.17}+...+\frac{4}{41.45}=-\frac{37}{45}\)

\(\Leftrightarrow x+3\left(\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+...+\frac{1}{41.45}\right)=-\frac{37}{45}\)

\(\Leftrightarrow x+\frac{3}{4}\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\right)=-\frac{37}{45}\)

\(\Leftrightarrow x+\frac{3}{4}\left(\frac{1}{5}-\frac{1}{45}\right)=-\frac{37}{45}\)

\(\Leftrightarrow x+\frac{3}{4}.\frac{8}{45}=-\frac{37}{45}\)

\(\Leftrightarrow x+\frac{2}{15}=-\frac{37}{45}\)

\(\Leftrightarrow x=-\frac{43}{45}\)

sai đề hay sao ý bạn

đề có vấn đề sao vậy bạn ?