A= 1.2² + 2.3² + 3.4² +.......+99.100²

A = 2².(2-1) + 3²(3-1) + 4².(4-1) +........+100².(100 -1)

A = 2³ - 2² + 3³ - 3² + 4³ - 4²+.......+100³ - 100²

A = (2³ + 3³ + 4³+..........+100³) - (2² + 3² + 4² +....+100²)

đặt C = 2² + 3² + 4²+......+100²

đặt B = 2³ + 3³ + 4³ + ....+100³ = 1³ + 2³ +3³ +4³+.....+100³ - 1³

áp dụng công thức tổng quát :

1³+ 2³ + 3³ + 4³ +....+ n³ = (1 + 2+3+...+n)²

ta có B = (1+2+3+.....+100)² - 1 = [(1+100).100:2]² - 1

        B =  5050² - 1

C = 2² + 3² + 4² + 5²+......+100²

C =2.(3-1) + 3(4-1)+........+100.(101-1)

C = 2.3 - 2 + 3.4 - 3 + .....+ 100.101 - 1

C = 2.3 + 3.4 + 4.5 + ....100.101 - ( 2 + 3 + 4 + 5+......+ 100)

C = 1.2+2.3+3.4+4.5+....+100.101-  1.2 -( 2 + 100)x {(100-2):1+1}:2

C = 100.101.102: 3 - 2 - 5049

C = 343400 - 5051

C = 338349

A = B - C = 5050² - 338349 

A = 25164151

6 cm A B C D

a)491-(x+83)=72

=> x+83=491-72

=> x+83=419

=> x = 336

b)6.(3x-1)+11=59

=> 6.(3x-1)=48

=> 3x-1=8

=>3x=9

=> x = 3

c) 15-(3x-1)=7

=>3x-1=8

=>3x-9

=>x-3

d)4x:25=0

=>4x=0

=>x=0

e)126-17.(3x-5)=7

=>17.(3x-5)=119

=>3x-5=7

=>3x=12

=>x=4

Đặt A = \(\dfrac{1}{5}+\dfrac{1}{45}+\dfrac{1}{117}+\dfrac{1}{221}+\dfrac{1}{357}+\dfrac{1}{525}+\dfrac{1}{725}+\dfrac{1}{957}\)

4A = \(\dfrac{4}{5}+\dfrac{4}{45}+\dfrac{4}{117}+\dfrac{4}{221}+\dfrac{4}{357}+\dfrac{4}{525}+\dfrac{4}{725}+\dfrac{4}{957}\)

4A = \(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+\dfrac{4}{13\cdot17}+\dfrac{4}{17\cdot21}+\dfrac{4}{21\cdot25}+\dfrac{4}{25\cdot29}+\dfrac{4}{29\cdot33}\)

4A = \(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{29}+\dfrac{1}{29}-\dfrac{1}{33}\)

4A = \(1-\dfrac{1}{33}\)

4A = \(\dfrac{32}{33}\)

A = \(\dfrac{32}{33}:4=\dfrac{8}{33}\)

d=(2007+2).(2011-2)=2007.2011-2.2007+2.2011-4=

=2007.2011+2(2011-2007)-4=2007.2011+4>c=2007.2011

\(C=2007.2011=\left(2009-2\right).\left(2009+2\right)\\ =2009^2-2.2=2009^2-4\\ D=2009.2009=2009^2\\ Vì:2009^2-4< 2009^2\\ \Rightarrow C< D\)

4³⁰⁰ = (4³)¹⁰⁰ = 64¹⁰⁰ = (32. 2)¹⁰⁰ = 32¹⁰⁰ .  2¹⁰⁰ 

12 . 24¹⁰⁰ < 16 . 24¹⁰⁰  = 2⁴ . 24¹⁰⁰ < 2⁴.32¹⁰⁰ < 2¹⁰⁰. 32¹⁰⁰ = 4³⁰⁰

vậy 4³⁰⁰ > 12. 24¹⁰⁰

\(4^{300}=\left(4^3\right)^{100}=64^{100}=\left(\dfrac{8}{3}\right)^{100}\times24^{100}\\ Vì:\left(\dfrac{8}{3}\right)^{100}>12\\ \Rightarrow4^{300}>12\times24^{100}\)

a) 2/3x-5/7=9/7

2/3x=9/7+5/7

2/3x=2

x=2:2/3

x=3

b)2/3x1/2021 + 2022/2021x2/3 - 2/3

2/3x (1/2021 + 2022/2021 - 1)

2/3 x 2/2021 = 4/6063

a, \(\dfrac{2}{3}\)x - \(\dfrac{5}{7}\) = \(\dfrac{9}{7}\)

\(\dfrac{2}{3}\) x = \(\dfrac{9}{7}\) + \(\dfrac{5}{7}\) 

\(\dfrac{2}{3}\)x = 2

   x = 2 : 2/3 

   x = 3

b, \(\dfrac{2}{3}\) x \(\dfrac{1}{2021}\) + \(\dfrac{2022}{2021}\) x \(\dfrac{2}{3}\) - \(\dfrac{2}{3}\)

= \(\dfrac{2}{3}\) ( \(\dfrac{1}{2001}\) + \(\dfrac{2022}{2021}\) - 1)

= \(\dfrac{2}{3}\) ( \(\dfrac{2023}{2021}\) - \(\dfrac{2021}{2021}\))

= \(\dfrac{2}{3}\) . \(\dfrac{2}{2021}\)

= \(\dfrac{4}{6063}\)

Ta có :

S = 3+2+4+8+...+2²⁰²²-2²⁰²³

=> S = 3+2+2²+2³+...+2²⁰²²-2²⁰²³

=> 2S=6+2²+2³+2⁴+...+2²⁰²³-2²⁰²⁴

=> 2S-S=(6+2²+2³+2⁴+...+2²⁰²³-2²⁰²⁴)-(3+2²+2³+...+2²⁰²²-2²⁰²³)

=> S = 3-2²⁰²⁴

S = 3 + 2 + 4 + 16 +......+ 2²⁰²² - 2²⁰²³

S =       (2 + 2² + 2³+......+2²⁰²²) + 3 - 2²⁰²³

đặt C = 2 + 2² + 2³+.......+2²⁰²² 

        C = 2 + 2² + 2³ +.........+ 2²⁰²²

      2C =       2² + 2³ + .......+  2²⁰²² + 2²⁰²³

2C - C = 2²⁰²³ - 2

         C = 2²⁰²³ - 2

   S = C + 3 - 2²⁰²³

   S =  2²⁰²³ - 2 + 3 - 2²⁰²³

    S = 1

`3(x+4)=2(x+15)`

`3x+12=2x+30`

`3x-2x=30-12`

`x=18`

   3.(x + 4) = 2.(x + 15)

⇒ 3x + 12 = 2x + 30

⇒  3x - 2x = 30 - 12

⇒           x = 18

`5( x + 2) + 4 ( x + 4) =125`

` 5x +10 + 4x +16 =125`

`x(5+4) = 125 -10 -16`

`9x = 99`

`x=99/9=11`

5(x+2)+4(x+4)=125

=> 5x+10+4x+16=125

=> 9x+26 = 125

=> 9x = 99

=> x = 11