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.

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

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

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

Ta có:

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

\(=\frac{7}{4}.\left[3333.\left(\frac{1}{1212}+\frac{1}{2020}+\frac{1}{3030}+\frac{1}{4242}\right)\right]\)

\(=\frac{7}{4}.\left[3333.\left(\frac{1}{12.101}+\frac{1}{20.101}+\frac{1}{30.101}+\frac{1}{42.101}\right)\right]\)

\(\frac{7}{4}.\left[3333.\frac{1}{101}\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\right]\)

\(=\frac{7}{4}.33.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)

\(=33.\left[\frac{7}{4}.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\right]\)

\(=33.\left[\frac{7}{4}.\left(\frac{1}{3}-\frac{1}{7}\right)\right]\)

\(=33.\left(\frac{7}{4}.\frac{4}{7}\right)\)

\(=33.1\)

\(=33\)

Vậy \(A=33\)

bạn làm sai rồi,1/3 - 1/7= 4/21 cơ mà. Và kết quả ra là 11

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

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

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

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

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

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

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

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! 

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

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

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

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

Rồi tách ra rồi làm nốt

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

\(=\dfrac{7}{4}\left(\dfrac{33\cdot101}{12\cdot101}+\dfrac{33\cdot101}{20\cdot101}+\dfrac{33\cdot101}{30\cdot101}+\dfrac{33\cdot101}{42\cdot101}\right)\)

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

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

\(=\dfrac{231}{4}\left(\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\right)\)

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

\(=\dfrac{231}{4}\left(\dfrac{1}{3}-\dfrac{1}{7}\right)=\dfrac{231}{4}\cdot\dfrac{4}{21}=11\)