\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{49\cdot50}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(=\dfrac{49}{50}< 1\)

Trả lời giúp mik hai phép tính còn lại nx nhé

Lời giải:

$A=(99-97)+(95-93)+(91-89)+...+(7-5)+(8-1)$

$=\underbrace{2+2+2+...+2}_{24}+7$

$=2\times 24+7$

$=55$

Ta có: 99–97 +95–93+...+7–5+3−1

=(99−97)+(95−93)+...+(7−5)+(3−1)

=2+2+...+2+2 (có 25 số 2)

=2.25

=50

mình muốn hỏi cách làm bạn ạ !!

Ta có 1/3x4= 1/3- 1/4

          2/4x6= 1/4-1/6

..............................

   5/13x18= 1/13 - 1/18

=> 1/3 -1/4 + 1/4 -1/6 +......+ 1/13 -1/18

  = 1/3 - 1/18

= 5/18

=1/3-1/4+1/4-1/6+1/6-1/9+...+1/13-1/18

=1/3-1/18

=5/18

k nha

2:

b=2000*2004

=(2002-2)*(2002+2)

=2002^2-4

=>b<a

1:

a: \(=8\cdot9\left(14+17+19\right)=72\cdot50=3600\)

Bài 1:

\(8\times9\times14+6\times17\times12+19\times4\times18\)

\(=8\times9\times14+3\times2\times17\times2\times2\times3+19\times4\times2\times9\)

\(=8\times9\times14+17\times8\times9+19\times8\times9\)

\(=8\times9\times\left(14+17+19\right)\)

\(=8\times9\times50\)

\(=72\times5\times10\)

\(=360\times10\)

\(=3600\)

Bài 2:

Ta có:

\(a=2022\times2022\)

Và: \(b=2000\times2004\)

Mà: \(2022>2000,2022>2004\)

\(\Rightarrow2022\times2022>2000\times2004\)

\(\Rightarrow a>b\)

=8,95*90+89,5

=89,5(9+1)

=895