K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

3 tháng 9 2017

\(F=\dfrac{49}{2.9}+\dfrac{49}{9.16}+............+\dfrac{49}{65.72}\)

\(\Leftrightarrow F=\dfrac{7^2}{2.9}+\dfrac{7^2}{9.16}+............+\dfrac{7^2}{65.72}\)

\(\Leftrightarrow F=7\left(\dfrac{7}{2.9}+\dfrac{7}{9.16}+.............+\dfrac{7}{65.72}\right)\)

\(\Leftrightarrow F=7\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...........+\dfrac{1}{65}-\dfrac{1}{75}\right)\)

\(\Leftrightarrow F=7\left(\dfrac{1}{2}-\dfrac{1}{72}\right)\)

\(\Leftrightarrow F=7.\dfrac{35}{72}=\dfrac{245}{72}\)

\(G=\dfrac{3}{1.3}+\dfrac{3}{3.5}+...........+\dfrac{3}{47.49}\)

\(\Leftrightarrow G=\dfrac{3.2}{1.3.2}+\dfrac{3.2}{3.5.2}+........+\dfrac{3.2}{47.49}\)

\(\Leftrightarrow G=\dfrac{3}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+..........+\dfrac{2}{47.49}\right)\)

\(\Leftrightarrow G=\dfrac{3}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+........+\dfrac{1}{47}-\dfrac{1}{49}\right)\)

\(\Leftrightarrow G=\dfrac{3}{2}\left(1-\dfrac{1}{49}\right)\)

\(\Leftrightarrow G=\dfrac{3}{2}.\dfrac{48}{49}=\dfrac{72}{49}\)

11 tháng 5 2022

\(B=\dfrac{49}{2\cdot9}+\dfrac{49}{9\cdot16}+\dfrac{49}{16\cdot23}+...+\dfrac{49}{65\cdot72}\)

\(B=\dfrac{49}{7}\cdot\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{23}+...+\dfrac{1}{65}-\dfrac{1}{72}\right)\)

\(B=7\cdot\left(\dfrac{1}{2}-\dfrac{1}{72}\right)\)

\(B=7\cdot\left(\dfrac{36}{72}-\dfrac{1}{72}\right)\)

\(B=7\cdot\dfrac{35}{72}\)

\(B=\dfrac{\left(7\cdot35\right)}{72}\)

\(B=\dfrac{245}{72}\)

11 tháng 5 2022

\(\dfrac{B}{7}=\dfrac{7}{2\cdot9}+\dfrac{7}{9\cdot16}+\dfrac{7}{16\cdot23}+...+\dfrac{49}{65\cdot72}\\ \dfrac{B}{7}=\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{23}+...+\dfrac{1}{65}-\dfrac{1}{72}\\ \dfrac{B}{7}=\dfrac{1}{2}-\dfrac{1}{72}\\ \dfrac{B}{7}=\dfrac{35}{72}\\ B=\dfrac{35}{72}\times7\\ B=\dfrac{245}{72} \)

=>2/1*3+2/3*5+...+2/(2x-1)(2x+1)=98/99

=>1-1/3+1/3-1/5+...+1/(2x-1)-1/(2x+1)=98/99

=>1-1/(2x+1)=98/99

=>1/(2x+1)=1/99

=>2x+1=99

=>x=49

16 tháng 7 2017

\(\frac{x}{1.3}+\frac{x}{3.5}+\frac{x}{5.7}+....+\frac{x}{97.99}=\frac{49}{99}\)

\(\Leftrightarrow\frac{x}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{97}-\frac{1}{99}\right)=\frac{49}{99}\)

\(\Leftrightarrow\frac{x}{2}.\left(\frac{1}{1}-\frac{1}{99}\right)=\frac{49}{99}\)

\(\Leftrightarrow\frac{x}{2}.\frac{98}{99}=\frac{49}{99}\)

\(\Leftrightarrow\frac{x}{2}=\frac{49}{99}\div\frac{98}{99}\)

\(\Leftrightarrow\frac{x}{2}=\frac{1}{2}\)

\(\Leftrightarrow x=\frac{1}{2}\times2=1\)

16 tháng 7 2017

\(\frac{x}{1\cdot3}+\frac{x}{3\cdot5}+...+\frac{x}{97\cdot99}=\frac{49}{99}\)

\(\Rightarrow\frac{x}{2}\left[\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{97\cdot99}\right]=\frac{49}{99}\)

\(\Rightarrow\frac{x}{2}\left[\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right]=\frac{97}{99}\)

\(\Rightarrow\frac{x}{2}\left[1-\frac{1}{99}\right]=\frac{49}{99}\)

\(\Rightarrow\frac{x}{2}\cdot\frac{98}{99}=\frac{49}{99}\)

\(\Rightarrow\frac{x}{2}=\frac{1}{2}\)

=> x = 1/2 * 2 = 1

20 tháng 4 2016

1/1-1/3+1/3-1/5+1/5-1/7+...... +1/47-1/49

20 tháng 4 2016

 3/1.3+3/3.5+3/5.7+......+3/47.49

=1/1-1/3+1/3-1/5+1/5-1/7+........+1/47-1/49

=1/1-1/49

=49/49-1/49

=48/49

20 tháng 8 2023

\(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{\left(2x-1\right)\cdot\left(2x+1\right)}=\dfrac{49}{99}\)

\(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{\left(2x-1\right)\cdot\left(2x+1\right)}=\dfrac{98}{99}\)

\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+....+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}=\dfrac{98}{99}\)

\(1-\dfrac{1}{2x+1}=\dfrac{98}{99}\)

\(\dfrac{2x+1-1}{2x+1}=\dfrac{98}{99}\)

\(\dfrac{2x}{2x+1}=\dfrac{98}{99}\)

=> 2x=98

=> x=49

22 tháng 3 2016

A = 7 (7 / 2.9 + 7 / 9.16 + .......... + 7/65.72)

A=7( 1/2 - 1/9 +1/9 - 1/16 +......+1/65 - 1/72)

A= 7 ( 1/2 -1/72)

A= 7 . 35/72

A=245/72

22 tháng 3 2016

\(A=\frac{7^2}{2.9}+\frac{7^2}{9.16}+\frac{7}{16.23}+.....+\frac{7^2}{65.72}\)

=\(7.\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+\frac{1}{16}-\frac{1}{23}+....+\frac{1}{65}-\frac{1}{72}\right)\)

=\(7.\left(\frac{1}{2}-\frac{1}{72}\right)\)

=\(7.\frac{35}{72}\)

=\(\frac{245}{72}\)

29 tháng 4 2018

\(\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left\{\left(2x+1\right).\left(2x+3\right)\right\}}\right)=\frac{49}{99}\)

\(\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}\right)=\frac{49}{99}\)

\(\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{2x+3}\right)=\frac{49}{99}\)

\(\frac{1}{2}.\cdot\left(\frac{2x+3}{2x+3}-\frac{1}{2x+3}\right)=\frac{49}{99}\)

\(\frac{1}{2}.\frac{2x+2}{2x+3}=\frac{49}{99}\)

\(\frac{2x+2}{2x+3}=\frac{49}{99}:\frac{1}{2}\)

\(\frac{2x+2}{2x+3}=\frac{98}{99}\)

=) \(2x+2=98\)và \(2x+3=99\)

TH1 : \(2x+2=98\)

\(2x=98-2\)

\(2x=96\)

\(x=96:2\)

\(x=48\)( THỎa mãn )

TH2 : 
\(2x+3=99\)

\(2x=99-3\)

\(2x=96\)

\(x=96:2\)

\(x=48\)( THỎa mãn )

Vậy x = 48

29 tháng 4 2018

Đặt A=