S = 2^4.(1^4+2^4+3^4+.....+10^4)

   = 16 . 25333

   = 405328

Tk mk nha

\(S=\left(1\cdot2\right)^4+\left(2\cdot2\right)^4+...+\left(10\cdot2\right)^4\)

\(S=1^4\cdot2^4+2^4\cdot2^4+...+10^4\cdot2^4\)

\(S=2^4\cdot\left(1^4+2^4+3^4+...+10^4\right)\)

\(S=16\cdot25333=405328\)

\(A=2^4+4^4+6^4+...+18^4+20^4\)

\(=2^4\left(1^4+2^4+3^4+...+9^4+10^4\right)\)

\(=16.25333=405328\)

mk k cho bn rồi đó

 k mk nha

\(A=2^4+4^4+...+18^4\)

\(=2^4.\left(1^4+2^4+...+10^4\right)\)

\(=16.25333\)

\(=405328\)

`1/2+2/4+3/6+4/8+5/10+6/12`

`=1/2+1/2+1/2+1/2+1/2+1/2`

`=1/2*6=3`

`1/3+1/4+1/5+8/10+20/15+20/30`

`=(1/3+1/4)+(1/5+4/5)+(4/3+2/3)`

`=7/12+1+2`

`=7/12+3=43/12`

\(\dfrac{1}{2}+\dfrac{2}{4}+\dfrac{3}{6}+\dfrac{4}{8}+\dfrac{5}{10}+\dfrac{6}{12}\)
\(=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}\)
\(=\dfrac{1}{2}\times6=3\)
\(------\)
\(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{8}{10}+\dfrac{20}{15}+\dfrac{20}{30}\)
\(=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{4}{5}+\dfrac{4}{3}+\dfrac{2}{3}\)
\(=\left(\dfrac{1}{3}+\dfrac{4}{3}+\dfrac{2}{3}\right)+\left(\dfrac{1}{5}+\dfrac{4}{5}\right)+\dfrac{1}{4}\)
\(=\dfrac{7}{3}+1+\dfrac{1}{4}\)
\(=\dfrac{28}{12}+\dfrac{12}{12}+\dfrac{3}{12}\)
\(=\dfrac{43}{12}\)

`=1/2+1/2+1/2+1/2+1/2+1/2= 1/2 \times 6=3`

`----`

`=1/3+1/4+1/5+4/5+4/3+2/3 =(4/3+2/3+)+(1/5+4/5)+1/3+1/4=2+1+1/3+1/4=3+1/3+1/4=43/12`

Bài 1:

A = 1 + 3 + 3² + ... + 3¹⁰⁰

=> 3A = 3 + 3² + ... + 3¹⁰¹

=> 2A = 3¹⁰¹ - 1

=> A = \(\frac{3^{101}-1}{2}\)

B = 1 + 4² + 4⁴ + ... + 4¹⁰⁰

=> 8B = 4² + 4⁴ + ... + 4¹⁰²

=> 7B = 4¹⁰² - 1

=> B = \(\frac{4^{102}-1}{7}\)

Bài 2:

a) S₁ = 2² + 4² + ... + 20²

=> S₁ = 2²(1+2²+...+10²)

=> S₁ = 2².385

=> S₁ = 1540

b) S₂ = 100² + 200² + ... + 1000²

=> S₂ = 100²(1+2²+...+10²)

=> S₂ = 100².385

=> S₂ = 3850000

(10+20)+(2+8)+(4+6)=30+10+10

=50

1/3+1/6+4/3+2/6+2/3+5/6

=(1/3+4/3+2/3)+(1/6+2/6+5/6)

=7/3+8/6=7/3+4/3=11/3

11/3

Chuyển 1⁴ thành 1. Và coi ... là x. Vậy:

\(1^4+1^4+x+9^4+10^4=\)\(25333\)

\(1+1+x+9^4+10^4=25333\)

\(\left(1+1\right)+x+9^4+10^4=\)\(25333\)

\(2+x+6561+10000=\)\(25333\)

\(x+\left(2+6561+10000\right)=\)\(25333\)

\(x+16563=25333\)

                     \(x=25333-16563\)

                     \(x=8770\)

Vậy số cần điền vào chỗ trống là 8770.

Học tốt nha.

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