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

Đặt A = 101 + 100 + 99 + 98 + ... + 1

Số số hạng của tổng A là :

(101 - 1) : 1 + 1 = 101 (số hạng)

Suy ra : A = (101 + 1) x 101 : 2 = 5151

Đặt B = 101 - 100 + 99 - 98 + ... + 3 - 2 + 1           (Mẫu số sai đề)

B = (101 - 100) + (99 - 98) + ... + (3 - 2) + 1    (Có : (101 - 3) : 2 + 1 = 50 cặp)

B = 1 + 1 + ... + 1 + 1          (Có : 50 + 1 = 51 số hạng 1)

B = 1 x 51

B = 51

Thay A,B vào (1), ta được :

\(\dfrac{101+100+99+98+...+1}{101-100+99-98+...+2-1}\) = \(\dfrac{5151}{51}\)= 101

* Mẫu số sai đề

ko hỉu

ai 2k11 ko

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

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

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

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 \(B=0\)

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

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

Ta có:

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

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

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

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

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

a) \(A=\dfrac{3737.43-4343.37}{2+4+6+...+2018}\)

\(\Leftrightarrow A=\dfrac{3737.43-43.101.37}{2+4+6+...+2018}\)

\(\Leftrightarrow A=\dfrac{3737.43-43.3737}{2+4+6+...+2018}\)

\(\Leftrightarrow A=\dfrac{0}{2+4+6+...+2018}\)

\(\Leftrightarrow A=0\)

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

\(\Leftrightarrow B=\dfrac{\left(101+1\right).101:2}{\left(101+99+97+...+1\right)-\left(100+98+96+...+2\right)}\)

\(\Leftrightarrow B=\dfrac{5151}{\left[\left(101+1\right).51:2\right]-\left[\left(100+2\right).50:2\right]}\)

\(\Leftrightarrow B=\dfrac{5151}{2601-2550}\)

\(\Leftrightarrow B=\dfrac{5151}{51}\)

\(\Leftrightarrow B=101\)