\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)

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

\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)

~ Hok tốt ~

\(\)

Viết thành 2 . (1/3.5 + 1/5.7 + 1/7.9 + ...+ 1/97.99

Tui hk bít nữa

ta co : 65%=0,65

goi A= 4.(1/3.5+1/5.7+1/7.9+............+1/97.99)

2A=4.( 2/3.5+2/5.7+2/7.9+...............+2/97.99)

2A=4.(1/3-1/5+1/5-1/7+1/7-1/9+...+1/97-1/99)

2A=4.(1/3-1/99)

2A=4.(33/=99+1/99)

2A=4.34/99

2A=136/99

A=136/99:2

A=68/99=0,69=0,68

Vi A=0,68 > 0,65

=> A > 65%

\(\dfrac{4}{1.3}+\dfrac{4}{3.5}+\dfrac{3}{5.7}+...+\dfrac{4}{97.99}\)
\(=2\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\right)\)
\(=2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(=2\left(1-\dfrac{1}{99}\right)\)
\(=2\cdot\dfrac{98}{99}\)
\(=\dfrac{196}{99}\)
#NoSimp

=2(2/1*3+2/3*5+...+2/97*99)

=2(1-1/3+1/3-1/5+...+1/97-1/99)

=2*98/99=196/99

\(\dfrac{4}{1.3}+\dfrac{4}{3.5}+\dfrac{4}{5.7}+...+\dfrac{4}{97.99}\)

=4.(\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\)

=4.1/2.(2/1.3+2/3.5+2/5.7+...+2/97.99)

=2.(1/1-1/3+1/3-1/5+1/5-1/7+...+1/97-1/99)

=2.(1/1-1/99)

=2.(99/99-1/99)

=2.98/99

=196/99

= \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\)

\(=\dfrac{1}{3}-\dfrac{1}{99}=\dfrac{32}{99}\)

= \(\dfrac{\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}}{2}\)

= \(\dfrac{\dfrac{1}{3}-\dfrac{1}{99}}{2}=\dfrac{16}{99}\)

Bài kia mk nhầm sorry