a: A=3^2(1^2+2^2+...+10^2)

=9*385

=3465

b: B=2^3(1^3+2^3+...+10^3)

=8*3025

=24200

We have : 1³ + 2³ + 3³ + ...... + 10³ = 3025

Then        2³( 1³ + 2³ + 3³ + ...... + 10³ ) = 3025.2³

<=>         2³ + 4³ + 6³ + ...... + 20³ = 3025.8

=>           2³ + 4³ + 6³ + ...... + 20³ = 24200

ai nhanh nhất mình k cho

ai nhanh nhất mình k cho

S=2^3+4^3+6^3+....+20^3=2^3×(1^3+2^3+3^3+....+10^3)=8×3025=24200

Có f=2³+4³+6³+...+20³

=>f=2³+2³.2³+2³.3³+...+2³.10³

=>f=2³(1³+2³+3³+...+10³)

=>f=8.3025=24200

Vậy f=24200

C=6050

lời giải

\(B=2^3+4^3+6^3+...+20^3\)

\(=2.\left(1^3+2^3+3^3+...+10^3\right)\)

\(=2.3025\)

\(=6050\)

Ta có B = 2³ + 4³ + 6³ + ... + 20³

             = 2³.(1³ + 2³ + 3³ + ... + 10³)

             = 2³.3025

             = 8. 3025

             = 24200

quên bài tôi sai rồi @@

a) Ta có: \(\dfrac{-5}{7}\left(\dfrac{14}{5}-\dfrac{7}{10}\right):\left|-\dfrac{2}{3}\right|-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)+\dfrac{10}{3}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)

\(=\dfrac{-5}{7}\cdot\dfrac{3}{2}\cdot\dfrac{21}{10}-\dfrac{3}{4}\cdot\dfrac{56}{3}+\dfrac{10}{3}\cdot\dfrac{8}{15}\)

\(=\dfrac{-9}{4}-14+\dfrac{16}{9}\)

\(=\dfrac{-1621}{126}\)

b) Ta có: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)

\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)

\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)

\(=-\dfrac{49}{20}\)

\(S=2^3+4^3+6^3+20^3\)

\(=2^3\cdot1^3+2^3\cdot2^3+2^3\cdot3^3+.....+2^3\cdot10^3\)

\(=2^3\left(1^3+2^3+3^3+....+10^3\right)\)

\(=8\cdot\left(1^3+2^3+3^3+...+10^3\right)\)

Mà \(1^3+2^3+3^3+....+10^3=3025\Rightarrow S=8\cdot3025=24200\)

Vậy S=24200