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

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

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

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

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

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

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

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

Đề là ntn:

\(A=49\left(\dfrac{1}{2.9}+\dfrac{1}{9.16}+\dfrac{1}{16.23}+...+\dfrac{1}{65.72}\right):\dfrac{1}{3}-\dfrac{7}{36}\)

\(A=7\left(\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{23}+...+\dfrac{1}{65}-\dfrac{1}{72}\right):\dfrac{1}{3}-\dfrac{7}{36}\)

\(A=7\left(\dfrac{1}{2}-\dfrac{1}{72}\right):\dfrac{1}{3}-\dfrac{7}{36}\)

\(A=7.\dfrac{35}{72}:\dfrac{1}{3}-\dfrac{7}{36}\)

\(A=\dfrac{245}{72}:\dfrac{1}{3}-\dfrac{7}{36}\)

\(A=\dfrac{735}{72}-\dfrac{7}{36}=\dfrac{735}{72}-\dfrac{14}{36}=\dfrac{721}{36}\)

em mới học lớp 7 thôi ạ

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

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

\(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x-1}-\frac{1}{2x+1}=\frac{98}{99}\)

\(\Rightarrow1-\frac{1}{2x+1}=\frac{98}{99}\)

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

=> 2x = 98

=> x = 98 : 2 = 49

cảm ơn bạn rất là nhiều nhé

ez

=>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

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

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

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

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

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

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

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

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

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

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

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

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

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

=> x = 1/2 * 2 = 1

nhân 2 vào 2 vế rồi bạn biến đổi ra( mình lười làm ắ)

tìm được x=50 ắ

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

 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

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

= 1 - 1/49

= 48/49 nha!

*** Ai k mk mk k lại !!***