Ta xét tử số 3+5+...+99+101

=[(101-3):2+1].[101+3):2]

=50.52

=2600 (1)

Ta xét mẫu số13.47+13.53

=13.(47+53)

=13.100

=1300   (2)

Thay (1) vả (2) vào biểu thức A ta được ;2600/1300=2

\(A=\frac{3+5+...+99+101}{13.47+13.53}\)

\(A=\left(3+5+...+99+101\right)\div\left(13.47+13.53\right)\)

\(A=\left\{101+3\cdot\left[\left(101-3\right)\cdot2+1\right]\div2\right\}\div\left[13.\left(47+53\right)\right]\)

\(A=2548\div1300\)

\(A=\frac{49}{25}\)

lộn \(A=2600\div1300=2\)

\(A=2\)

thông cảm nhé

cách tính đúng nhưng mà tớ ghi lộn kết quả

\(A=\dfrac{101\cdot\dfrac{102}{2}}{\left(101-100\right)+99-98+...+3-2+1}\)

\(=\dfrac{101\cdot51}{1+1+...+1}=\dfrac{101\cdot51}{51}=101\)

\(B=\dfrac{37\cdot43\left(101-101\right)}{2+4+...+100}=0\)

a, \(A=\dfrac{101+100+99+98+...+3+2+1}{101-100+99-98+...+3-2+1}\) 

Ta có: \(T=101+100+99+98+...+3+2+1\) \(=\dfrac{\left(101+1\right).101}{2}\) 

                                                                             \(=\dfrac{102.101}{2}\Leftrightarrow51.101\) 

  \(M=101-100+99-98+...+3-2+1\) 

Ta có: \(101:2=50\) (dư \(1\)) 

\(\Rightarrow M=\left(101-100\right)+\left(99-98\right)+...+\left(3-2\right)+1\) 

 Có \(50\) dấu ngoặc tròn "\(\left(\right)\)"

 \(\Rightarrow M=1+1+...+1+1=51.1=51\) 

      \(M\) có  \(51\) số \(1\)  

 \(\Rightarrow A=\dfrac{T}{M}=\dfrac{51.101}{51}=101\)

 Vậy \(A=101\)

b, \(B=\dfrac{3737.43-4343.37}{2+4+6+...100}\) 

Ta có: \(T=3737.43-4343.37\) 

          \(T=37.101.43-43.101.37\) 

          \(T=0\) 

\(\Rightarrow\) \(B=\dfrac{T}{2+4+6+...+100}=\dfrac{0}{2+4+6+...+100}\) \(=0\) 

 Vậy \(B=0\)

\(A=\dfrac{101+100+99+98+...+3+2+1}{101-100+99-98+...+3-2+1}.\)

\(A=\dfrac{\left[\dfrac{\left(101-1\right)}{1}+1\right]\left[\dfrac{101+1}{2}\right]}{\left(101-100\right)+\left(99-98\right)+...+\left(3-2\right)+1}.\)

\(A=\dfrac{101.51}{1+1+1+...+1+1}\) (có 51 số 1).

\(A=\dfrac{5151}{51}=101.\)

Vậy \(A=101.\)

Ta có:

A = \(\dfrac{101+100+99+98+...+1}{101-100+99-98+...+3-2+1}\)

= \(\dfrac{101+\left(100+1\right)+\left(99+2\right)+...+\left(51+50\right)}{\left(101-100\right)+\left(99-98\right)+...+\left(3-2\right)+1}\)

= \(\dfrac{101+101+101+...+101}{1+1+1+...+1}\) (51 số 101 và 51 số 1)

= \(\dfrac{101.51}{51}\)

= 101

Vậy A = 101

Gọi \(101+100+99+98+...+3+2+1\) là \(A\)

Gọi \(101-100+99-98+...+3-2+1\) là \(B\)

Ta có:

\(A=1+2+3+...+98+99+100+101\\ =\dfrac{101\cdot\left(101+1\right)}{2}\\ =\dfrac{101\cdot102}{2}\\ =5151\)

\(B=101-100+99-98+...+3-2+1\\ =\left(101-100\right)+\left(99-98\right)+...+\left(3-2\right)+1\\ =1+1+...+1+1\)

Ta có: A=1.2.3.....99.100.(\(1+\dfrac{1}{2}+\dfrac{1}{3}+......+\dfrac{1}{99}+\dfrac{1}{100}\))

      \(=1.2.3...100\left[\left(1+\dfrac{1}{100}\right)+\left(\dfrac{1}{2}+\dfrac{1}{99}\right)+......+\left(\dfrac{1}{50}+\dfrac{1}{51}\right)\right]\)

      =>A= 1.2...100.\(\left[\dfrac{101}{100}+\dfrac{101}{2.99}+......+\dfrac{101}{50.51}\right]\)

       =1.2.....100.101\(\left[\dfrac{1}{100}+\dfrac{1}{2.99}+.....+\dfrac{1}{50.51}\right]⋮101\)

               Vậy A chia hết cho 101