Vào fx ghi đi

A= 7/4 (33/12 +33/20+33/30+33/42)

A= 7/4 * 33 * (1/12+1/20+1/30+1/42)

A= 7/4 * 33 * (1/3*4 + 1/4*5 +1/5*6 +1/6*7)

A= 231/4 (1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7)

A=231/4(1/3-1/7)

A=231/4(7/21-3/21)

A=231/4 * 4/21

A=11

Vậy A=11

Ta có A=7/4.(3333/1212+3333/2020+3333/3030+3333/4242)

=7/4.(11/4+33/20+11/10+11/14

=7/4.44/7

=11

=> A=\(\frac{7}{4}\) . ( \(\frac{33}{12}\) + \(\frac{33}{20}\) + \(\frac{33}{30}\) + \(\frac{33}{42}\) ) => A=   \(\frac{7}{4}\).33. ( \(\frac{1}{12}\) + \(\frac{1}{20}\) + \(\frac{1}{30}\) + \(\frac{1}{42}\) )

=> A=\(\frac{7}{4}\).33. ( \(\frac{1}{3.4}\) + \(\frac{1}{4.5}\) + \(\frac{1}{5.6}\) + \(\frac{1}{6.7}\) ) = \(\frac{7}{4}\).33.(\(\frac{1}{3}\) - \(\frac{1}{4}\) + \(\frac{1}{4}\) - \(\frac{1}{5}\) + \(\frac{1}{5}\) - \(\frac{1}{6}\) + \(\frac{1}{6}\) - \(\frac{1}{7}\) )

=   \(\frac{7}{4}\) .33.(\(\frac{1}{3}\) - \(\frac{1}{7}\)) =  \(\frac{7}{4}\) .33. \(\frac{4}{21}\) = 11. Vậy A=11

Ta có:

\(\Rightarrow A=\frac{7}{4}.\left(\frac{11}{4}+\frac{33}{20}+\frac{11}{10}+\frac{11}{14}\right)\)

\(\Rightarrow A=\frac{7}{4}.\frac{44}{7}=11\)

\(A=\frac{7}{4}\cdot\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)

\(A=\frac{7}{4}\cdot\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)

\(A=\frac{7}{4}\cdot33\cdot\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)

Rồi tách ra rồi làm nốt

A = 7/4.[3333.(1/1212+1/2020+1/3030+1/4242)]

A= 7/4.[3333.(1/12.101+1/20.101+1/30.101+1/42.101)]

A= 7/4.[3333.(1/12+1/20+1/30+1/42)]

A= 7/4.[3333.(1/3.4+1/4.5+1/5.6+1/6.7)]

A= 7/4.[3333.(1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7)]

A= 7/4.[3333.(1/3-1/7)]

A= 7/4.[3333.4/21]

A= 7/4.4444/7

A=1111

\(A=\frac{7}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)

\(A=\frac{7}{4}.\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)

\(A=\frac{7}{4}.33.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)

\(A=\frac{231}{4}.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)

\(A=\frac{231}{4}.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)

\(A=\frac{231}{4}.\left(\frac{1}{3}-\frac{1}{7}\right)\)

\(A=\frac{231}{4}.\frac{4}{21}\)
\(A=11\)

A = 11 nha !
 

A= 7/4 . ( 3333/1212 + 3333/2020 + 3333/3030 + 3333/4242 ) 

  = 7/4 . ( 11/4 + 33/20 + 11/10 + 11/14 ) 

 = 7/4 . 44/7 

= 11  

Ps : Cộng phân số trong ngoặc bạn tự tính nha! 

A = 7/4 * ( 3333/1212 + 3333/2020 + 3333/3030 + 3333/4242)                                  => A = 7/4* (33/12 + 33/20 + 33/30 + 33/42)                                                              => A = 7/4* ( 33/3*4 + 33/4*5 + 33/5*6 + 33/6*7)                                                        => A = 7/4* { 33/(4-3) * ( 1/4 - 1/5 + 1/5 - 1/6 + 1/6-1/7)}                                              => A = 7/4*33 * ( 1/4 - 1/7)                                                                                                A = 231/4 * 3/28 =693/112.

\(\frac{7}{4}\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)=\frac{7}{4}\left(\frac{33\times101}{12\times101}+\frac{33\times101}{20\times101}+\frac{33\times101}{30\times101}+\frac{33\times101}{42\times101}\right)\)

\(=\frac{7}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)=\frac{7}{4}\times\frac{44}{7}=11\)

1/ \(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{8^2}\)

\(B< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{7.8}\)

\(B< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}\)

\(B< \dfrac{1}{1}-\dfrac{1}{8}< 1\)

\(B< 1\)

2/ \(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{20}\right)\)

\(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{19}{20}\)

\(B=\dfrac{1\times2\times3\times...\times19}{2\times3\times4\times...\times20}\)

\(B=\dfrac{1}{20}\)

3/ \(A=\dfrac{7}{4}\cdot\left(\dfrac{3333}{1212}+\dfrac{3333}{2020}+\dfrac{3333}{3030}+\dfrac{3333}{4242}\right)\)

\(A=\dfrac{7}{4}\cdot\left(\dfrac{33}{12}+\dfrac{33}{20}+\dfrac{33}{30}+\dfrac{33}{42}\right)\)

\(A=\dfrac{7}{4}\cdot\left(\dfrac{33}{3.4}+\dfrac{33}{4.5}+\dfrac{33}{5.6}+\dfrac{33}{6.7}\right)\)

\(A=\dfrac{7}{4}.33.\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)\)

\(A=\dfrac{231}{4}.\left(\dfrac{1}{3}-\dfrac{1}{7}\right)\)

\(A=\dfrac{231}{4}\cdot\dfrac{4}{21}\)

\(A=11\)

4/ A phải là \(\dfrac{2011+2012}{2012+2013}\)

Ta có : \(B=\dfrac{2011}{2012}+\dfrac{2012}{2013}>\dfrac{2011}{2013}+\dfrac{2012}{2013}=\dfrac{2011+2012}{2013}>\dfrac{2011+2012}{2012+2013}=A\)

\(\Rightarrow B>A\)