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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

Mình cảm ơn bạn nhiều

a: Đặt \(A=2+2^2+\cdots+2^{60}\)

Ta có: \(A=2+2^2+\cdots+2^{60}\)

\(=\left(2+2^2\right)+\left(2^3+2^4\right)+\cdots+\left(2^{59}+2^{60}\right)\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+\cdots+2^{59}\left(1+2\right)\)

\(=3\left(2+2^3+\cdots+2^{59}\right)\) ⋮3

Ta có: \(A=2+2^2+\cdots+2^{60}\)

\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+\cdots+\left(2^{58}+2^{59}+2^{60}\right)\)

\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+\cdots+2^{58}\left(1+2+2^2\right)\)

\(=7\left(2+2^4+\cdots+2^{58}\right)\) ⋮7

TA có: \(A=2+2^2+\cdots+2^{60}\)

\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+\ldots+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)

\(=2\left(1+2+2_{}^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+\cdots+2^{57}\left(1+2+2^2+2^3\right)\)

\(=15\left(1+2^5+\cdots+2^{57}\right)\) ⋮15

b: Ta có: \(B=1+3+3^2+\cdots+3^{1991}\)

\(=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+\cdots+\left(3^{1989}+3^{1990}+3^{1991}\right)\)

\(=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+\cdots+3^{1989}\left(1+3+3^2\right)\)

\(=13\left(1+3^3+\cdots+3^{1989}\right)\) ⋮13

c: Ta có: \(C=3+3^2+3^3+\cdots+3^{1998}\)

\(=\left(3+3^2\right)+\left(3^3+3^4\right)+\cdots+\left(3^{1997}+3^{1998}\right)\)

\(=\left(3+3^2\right)+3^2\left(3+3^2\right)+\cdots+3^{1996}\left(3+3^2\right)\)

\(=12\left(1+3^2+\cdots+3^{1996}\right)\) ⋮12

Ta có: \(C=3+3^2+3^3+\cdots+3^{1998}\)

\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+\cdots+\left(3^{1996}+3^{1997}+3^{1998}\right)\)

\(=\left(3+3^2+3^3\right)+3^3\left(3+3^2+3^3\right)+\cdots+3^{1995}\left(3+3^2+3^3\right)\)

\(=39\left(1+3^3+\cdots+3^{1995}\right)\) ⋮39

A= 1 - 2 - 2² - 2³ + 2⁴ +...+ 2²⁰²² (sửa đề)

 = -13 + (2⁴ + 2⁵ + 2⁶ + ... + 2²⁰²²)

2A = -26 + (2⁵ + 2⁶ + 2⁷ +  ... + 2²⁰²³)

2A - A = -26 + (2⁵ + 2⁶ + 2⁷ +  ... + 2²⁰²³) -  [-13 + (2⁴ + 2⁵ + 2⁶ + ... + 2²⁰²²)]

A = -13 +(2²⁰²³ - 2⁴)

= 2²⁰²³ - 29

Vậy...

B = 1 + 3 + 3² + 3³ + 3⁴ + ... + 3²⁰²²

3B = 3 + 3² + 3³ + 3⁴ + 3⁵ +...+ 3²⁰²³

3B - B = 3 + 3² + 3³ + 3⁴ + 3⁵ +...+ 3²⁰²³ - (1 + 3 + 3² + 3³ + 3⁴ + ... + 3²⁰²²)

2B = 3²⁰²³ - 1

=> B = \(\dfrac{3^{2023}-1}{2}\)

Vậy...

#Ayumu

A = 1/2 + 1/2^2 + 1/2^3 + ... + 1/2^n < 1

A = 1/2 + 1/2^2 + 1/2^3 + ... + 1/2^n

2A = 1 + 1/2 + 1/2^2+ ..+ 1/2^n-1

2A - A = 1 + 1/2 + 1/2^2+ ..+ 1/2^n-1 - (1/2 + 1/2^2 + 1/2^3 + ... + 1/2^n)

A = (1 - 1/2^n) + (1/2 - 1/2) + ..+ (1/2^n-1 -1/2^n-1)

A = 1 - 1/2^n

A < 1 (đpcm)

Câu b:

B = 1/3 + 1/3^2 + 1/3^3 + ...+ 1/3^n < 1/2

3B = 1 + 1/3 + 1/3^2 + ..+ 1/3^n - 1

3B - B = 1 + 1/3 + 1/3^2 + ..+ 1/3^n - 1 - (1/3 + 1/3^2 + 1/3^3 + ...+ 1/3^n)

2B = 1 + 1/3 + 1/3^2 + ..+ 1/3^n - 1 - 1/3 - 1/3^2 - 1/3^3 -..- 1/3^n-1 - 1/3^n

2B = (1 - 1/3^n) + (1/3 - 1/3) +..+(1/3^n-1-1/3^n-1)

2B = 1 - 1/3^n

B = 1/2 - 1/2.3^n < 1/2 (đpcm)

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

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

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

\(=-7+\dfrac{15-3}{10}=-7+\dfrac{6}{5}=-\dfrac{29}{5}\)

6: \(=6-\dfrac{2}{3}+\dfrac{1}{2}-5-\dfrac{5}{3}+\dfrac{3}{2}-3+\dfrac{7}{3}-\dfrac{5}{2}\)

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

\(=-2-\dfrac{1}{2}=-\dfrac{5}{2}\)

7: \(=\dfrac{5}{3}-\dfrac{3}{7}+9-2-\dfrac{5}{7}+\dfrac{2}{3}+\dfrac{8}{7}-\dfrac{4}{3}-10\)

\(=9-2-10+\dfrac{5}{3}+\dfrac{2}{3}-\dfrac{4}{3}+\dfrac{-3}{7}-\dfrac{5}{7}+\dfrac{8}{7}\)

=-3+1

=-2

8: \(=8-\dfrac{9}{4}+\dfrac{2}{7}+6+\dfrac{3}{7}-\dfrac{5}{4}-3-\dfrac{2}{4}+\dfrac{9}{7}\)

\(=8+6-3+\dfrac{2}{7}+\dfrac{3}{7}+\dfrac{9}{7}-1-\dfrac{2}{4}\)

\(=11+2-1-\dfrac{1}{2}\)

=11+1/2

=11,5

...

a) \(x-\dfrac{3}{4}=6\times\dfrac{3}{8}\)

\(x-\dfrac{3}{4}=\dfrac{9}{4}\)

=> \(x=\dfrac{9}{4}+\dfrac{3}{4}=3\)

b) \(\dfrac{7}{8}:x=3-\dfrac{1}{2}\)

\(\dfrac{7}{8}:x=\dfrac{5}{2}\)

=> \(x=\dfrac{7}{8}:\dfrac{5}{2}=\dfrac{7}{20}\)

c) \(x+\dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{3}{4}\)

\(x+\dfrac{1}{6}=\dfrac{3}{4}\)

=> \(x=\dfrac{3}{4}-\dfrac{1}{6}=\dfrac{7}{12}\)

d) \(\dfrac{3}{2}\times\dfrac{4}{5}-x=\dfrac{2}{3}\)

\(\dfrac{6}{5}-x=\dfrac{2}{3}\)

=> \(x=\dfrac{6}{5}-\dfrac{2}{3}=\dfrac{8}{15}\)

e) \(x\times3\dfrac{1}{3}=3\dfrac{1}{3}:4\dfrac{1}{4}\)(?)

\(x\times\dfrac{10}{3}=\dfrac{40}{51}\)

=> \(x=\dfrac{40}{51}:\dfrac{10}{3}=\dfrac{4}{17}\)

f) \(5\dfrac{2}{3}:x=3\dfrac{2}{3}-2\)

\(\dfrac{17}{3}:x=\dfrac{5}{3}\)

=> \(x=\dfrac{17}{3}:\dfrac{5}{3}=\dfrac{17}{5}\)

a: =>x-3/4=18/8=9/4

=>x=9/4+3/4=12/4=3

b: =>7/8:x=5/2

=>x=7/8:5/2=7/8*2/5=14/40=7/20

c: x+1/2*1/3=3/4

=>x+1/6=3/4

=>x=3/4-1/6=9/12-2/12=7/12

d: =>12/10-x=2/3

=>6/5-x=2/3

=>x=6/5-2/3=18/15-10/15=8/15

e: =>x*10/3=10/3:17/4=10/3*4/17

=>x=4/17

f: =>17/3:x=13/3-5/2=26/6-15/6=11/6

=>x=17/3:11/6=17/3*6/11=34/11