1/1.2 + 1/2.3 + .................+ 1/99.100 =

1/1 - 1/2 + 1/2 - 1/3 +....................+ 1/99 - 1/100 =

1/1 - 1/100                                                         =   99/100

98.99/99.100

99/100

`1/( 1.2 ) + 1/( 2.3 ) + .......+1/(99.100)`

`= 1-1/2+1/2-1/3+.....+1/99-1/100`

`=1-1/100`

`=99/100`

=1-1/2+1/2-1/3+...+1/99-1/100

=1-1/100=99/100

\(=\dfrac{1}{1.2}+\dfrac{1}{2.3}+....+\dfrac{1}{99.100}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)

Ta có : 1.98 + 2.97 + 3.96 + ...+ 98.1 = 1 + ( 1 + 2 ) + ( 1 + 2 + 3 ) + .....+ ( 1 + 2 + 3 + ...+ 97 + 98 ) = \(\frac{1.2}{2}\)+ \(\frac{2.3}{2}\)+  \(\frac{3.4}{2}\)+ ...+ \(\frac{98.99}{2}\)= \(\frac{1}{2}\)( 1 . 2 + 2 . 3 + 3 . 4 +...+ 98 . 99).

Vậy A = \(\frac{1}{2}\)

Nè bạn giải cụ thể chi tiết cho mình đk k thì mình mới k cho đk

2.
a) (2x + 1)³ = 125
    (2x + 1)³ = 5³
     2x + 1    = 5
     2x          = 5 - 1
     2x          = 4
       x          = 4:2
       x          = 2
Vậy x = 2
b) 5^x+1 = 5⁴
    x + 1 = 4
    x       = 4 - 1 
    x       = 3
Vây x = 3

a) \(A=1.2+2.3+3.4+...+98.99+99.100\)

\(3A=1.2.3+2.3.4+3.4.3+...+99.100.3\) 

\(3A=1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)....99.100.\left(101-98\right)\) 

\(3A=\left(1.2.3+2.3.4+3.4.5+...+99.100.101\right)-\left(0.1.2+1.2.3+2.3.4+...+98.99.100\right)\)

\(3A=99.100.101-0.1.2\) 

\(3A=999900-0\)

\(3A=999900\)

\(A=999900:3\)

\(\Rightarrow A=333300\)

a) \(\left(2x+1\right)^3=125\)

\(\left(2x+1\right)^3=5^3\)

\(2x+1=5\)

\(2x=5-1\) 

\(2x=4\)

\(x=4:2\)

\(\Rightarrow x=2\)

b) \(5^{x+1}=625\)

\(5^{x+1}=5^4\)

\(x+1=4\)

\(x=4-1\)

\(\Rightarrow x=3\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(A=1-\frac{1}{100}=\frac{99}{100}\)

vì \(\frac{99}{100}< 1\)

nên \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}< 1\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)

\(A=1-\frac{1}{100}< 1\)

Vậy A<1

Ta có: \(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)

=>\(A=1-\frac{1}{100}\)

Vì \(\frac{1}{100}>0\Rightarrow\)\(1-\frac{1}{100}< 1\)hay A<1

adct: (n(n+1)(n+2))/3 =>((98(98+1)(98+2))/3=323400

Đặt \(A=1.2+2.3+...+98.99\)

\(3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+....+98.99.\left(100-97\right)\)

\(3A=1.2.3+2.3.4-1.2.3+...+98.99.100-97.98.99\)

\(3A=98.99.100\Rightarrow A=\frac{98.99.100}{3}\)

\(\Rightarrow\frac{\frac{98.99.100}{3}.x}{26950}=\frac{-60}{7}\)\(\Rightarrow98.99.100.x=-\frac{60}{7}.80850\)

\(\Rightarrow98.99.100.x=-693000\)

Đến đây bạn tự tính nhé

cảm ơn bạn nhé, nếu không có bạn chắc mai mình bị cô mắng chết

\(B=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}\)

\(=\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{99-98}{98.99}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}\)

\(=1-\dfrac{1}{99}\)

\(A=\dfrac{2021}{2022}=\dfrac{2022-1}{2022}=1-\dfrac{1}{2022}\)

Có \(2022>99>0\Leftrightarrow\dfrac{1}{99}>\dfrac{1}{2022}\)

Suy ra \(A>B\).