\(\dfrac{2x}{15}+\dfrac{2x}{35}+\dfrac{2x}{63}+...+\dfrac{2x}{195}=\dfrac{4}{5}\\ x\cdot\left(\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+...+\dfrac{2}{195}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{13\cdot15}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{13}-\dfrac{1}{15}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{1}{3}-\dfrac{1}{15}\right)=\dfrac{4}{5}\\ x\cdot\dfrac{4}{15}=\dfrac{4}{5}\\ x=\dfrac{4}{5}:\dfrac{4}{15}\\ x=3\)

Gọi \(D=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\)

\(2D=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\\ 2D+D=\left(1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\right)\\ 3D=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\\ 3D=1-\dfrac{1}{64}< 1\\ \Rightarrow D=\dfrac{1-\dfrac{1}{64}}{3}< \dfrac{1}{3}\)

Vậy \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}< \dfrac{1}{3}\)

a) 3/7 + 4/9 + 4/7 + 5/9

= ( 3/7 + 4/7 ) + ( 4/9 + 5/9 )

= 7/7 + 9/9

= 1  + 1 

= 2

b)1/5 + 4/10 + 9/15 + 16/20 + 25/25 + 36/30 + 49/35 + 64/40 + 81/45

= 1/5 + 2/5 + 3/5 + 4/5 + 5/5 + 6/5 + 7/5 + 8/5 + 9/5

= ( 1/5 + 9/5 ) + ( 2/5 + 8/5 ) + (7/5 + 3/5 ) + ( 4/5 + 6/5  ) + 5/5

= 2 + 2 + 2 + 2 + 1

= 2  x 4 + 1

= 8  +1 

= 9

c)  1/8 + 1/12 + 3/8 + 5/12

= ( 1/8 + 3/8 ) + ( 1/12 + 5/12)

= 4/8 + 6/12

= 1/2 + 1/2

= 2/4 = 1/2

mỏi tay rồi

d; (1 - \(\dfrac{1}{2}\)) x (1 - \(\dfrac{1}{3}\)) x (1 - \(\dfrac{1}{4}\)) x ... x ( 1 - \(\dfrac{1}{100}\))

 = \(\dfrac{1}{2}\) x \(\dfrac{2}{3}\)  x \(\dfrac{3}{4}\) x \(\dfrac{3}{4}\) x ... x \(\dfrac{99}{100}\)

= \(\dfrac{1}{100}\)

e; \(\dfrac{9}{5}\) : \(\dfrac{17}{15}\) + \(\dfrac{8}{5}\) : \(\dfrac{17}{15}\)

=  \(\dfrac{9}{5}\) x \(\dfrac{15}{17}\) + \(\dfrac{8}{5}\) x \(\dfrac{15}{17}\)

=  \(\dfrac{27}{17}\) + \(\dfrac{24}{17}\)

=  \(\dfrac{51}{17}\)

=   3

a)\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{195}\)

Đặt \(C=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{66}\)

\(\Rightarrow\frac{1}{2}C=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{132}\)

\(\Rightarrow\frac{1}{2}C=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{11.12}\)

\(\Rightarrow\frac{1}{2}C=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}\)

\(\Rightarrow\frac{1}{2}C=\frac{1}{4}+\left(-\frac{1}{5}+\frac{1}{5}\right)+\left(-\frac{1}{6}+\frac{1}{6}\right)+...+\left(-\frac{1}{11}+\frac{1}{11}\right)-\frac{1}{12}\)\(\Rightarrow\frac{1}{2}C=\frac{1}{4}+0+0+...+0-\frac{1}{12}\)

\(\Rightarrow\frac{1}{2}C=\frac{1}{4}-\frac{1}{12}\)

\(\Rightarrow\frac{1}{2}C=\frac{3}{12}-\frac{1}{12}\)

\(\Rightarrow\frac{1}{2}C=\frac{2}{12}\)

\(\Rightarrow\frac{1}{2}C=\frac{1}{6}\)

\(\Rightarrow C=\frac{1}{6}:\frac{1}{2}\)

\(\Rightarrow C=\frac{1}{6}\cdot2\)

\(\Rightarrow C=\frac{2}{6}=\frac{1}{3}\)

\(\frac{1}{5}+\frac{4}{10}+\frac{9}{15}+\frac{16}{20}+1+\frac{36}{30}+\frac{49}{35}+\frac{64}{40}+\frac{81}{45}\)

\(=\left(\frac{1}{5}+\frac{81}{45}\right)+\left(\frac{4}{10}+\frac{49}{35}\right)+\left(\frac{9}{15}+\frac{49}{35}\right)+\left(\frac{16}{20}+\frac{36}{30}\right)+1\)

\(=2+2+2+2+1\)

\(=2\times4+1\)

\(=9\)

~ Hok tốt ~

a)\(3-\left(\frac{1}{4}+\frac{2}{3}\right)-\left(5+\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)

=\(3-\frac{1}{4}-\frac{2}{3}-5-\frac{1}{3}+\frac{6}{5}-6+\frac{7}{4}-\frac{3}{2}\)

=\(\left(3-5-6\right)+\left(\frac{-1}{4}+\frac{7}{4}\right)+\left(\frac{-2}{3}-\frac{1}{3}\right)+\left(\frac{6}{5}-\frac{3}{2}\right)\)

=\(-8+\frac{3}{2}-1-\frac{3}{10}\)

=\(\left(-8-1\right)+\left(\frac{3}{2}-\frac{3}{10}\right)\)

=-9+\(\frac{6}{5}\)

=\(\frac{-39}{5}\)