ta có 

3737 . 43 - 4343 . 37 = ( 3700 + 37) . 43 - (4300+43) . 37

= 3700 . 43 + 37 .43 - 4300 . 37 + 43 . 37

= 43 . 37 . ( 3700 . 43 - 4300 . 37)

= 1591 . (159100 - 159100)

= 1591 . 0

= 0

vậy \(\frac{3737.43-4343.37}{2+4+6+....+100}\)= 0

đap số 0

1 nha bạn

Chúc các bạn học giỏi

Tết vui vẻ nha

bằng 0 nhé =))))

\(b=\frac{3737\cdot43-4343\cdot37}{2+4+6+...+100}=\frac{37\cdot101\cdot43-43\cdot101\cdot37}{2+4+6+...+100}=\frac{0}{2+4+6+...+100}=0\)

\(B=\frac{3737\cdot43-4343\cdot37}{2+4+6+...+100}=\frac{37\cdot101\cdot43-43\cdot101\cdot37}{2+4+6+...+100}=\frac{0}{2+4+6+...+100}=0\)

`(3737.43-4343.37)/(2+4+...+100)`

`= (101.37.43-101.43.37)/(2+4+...+100)`

`=0/(2+4+...+100) = 0`

\(\dfrac{3737.43-4343.37}{2+4+6+...+100}\) 

Gọi \(A=2+4+6+...+100\) 

Số số hạng: \(\left(100-2\right):2+1=50\) 

Tổng dạy: \(\left(2+100\right).50:2=2550\) 

Gọi \(B=3737.43-4343.37\) 

\(B=101.37.43-101.43.37=0\) 

\(\Rightarrow\dfrac{3737.43-4343.37}{2+4+6+...+100}=\dfrac{0}{2550}=0\)

\(D=\frac{3737.43-4343.37}{2+4+6+...+100}\)

\(\Rightarrow D=\frac{37.101.43-4343.37}{2+4+6+...+100}\)

\(\Rightarrow D=\frac{37.4343-4343.37}{2+4+6+...+100}\)

\(\Rightarrow D=\frac{4343.\left(37-37\right)}{2+4+6+...+100}\)

\(\Rightarrow D=\frac{4343.0}{2+4+6+...+100}\)

\(\Rightarrow D=\frac{0}{2+4+6+...+100}\)

\(\Rightarrow D=0\)

Vậy D = 0

D=0

ngắn gọn

\(D=\frac{3737.43-4343.37}{2+4+6+...+100}\)

Biến đổi tử số ta có :

\(D_1=3737.43-4343.37\)

\(\Rightarrow D_1=37.101.43-4343.37\)

\(\Rightarrow D_1=\left(43.101\right).37-4343.37\)

\(\Rightarrow D_1=4343.37-4343.37\)

\(\Rightarrow D_1=4343\left(37-37\right)=4343.0=0\)

Thay vào , ta có :

\(\frac{0}{2+4+6+...+100}=0\div\left(2+4+6+...+100\right)=0\)

Vậy : D = 0

\(D=\frac{37.101.43-43.101.37}{2+4+6+..+100}=\frac{0}{2+4+...+100}=0\)

C=\(\frac{101+100+...+3+2+1}{101-100+...+3-2+1}\)

=\(\frac{\left(101+1\right).101:2}{\left(101-100\right)+...+\left(3-2\right)+1}\) (nhóm 2 số hạng ở MS thì sẽ có 51 nhóm và dư 1 số hang )

=\(\frac{102.101:2}{1+...+1+1}\) ( Ms có 51 số 1)

=\(\frac{51.101}{51}\)=101

D=\(\frac{3737.43-4343.37}{2+4+6+...+100}\)

= \(\frac{37.101.43-43.101.37}{2+4+6+..+100}\)

= \(\frac{0}{2+4+6+...+100}\)

=0

Tick mik nha, thks bạn

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

\(=\dfrac{37.101.43-4343.37}{2+4+6+...+100}\)

\(=\dfrac{37.4343-4343.37}{2+4+6+...+100}\)

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

\(=0\)

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

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

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

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 \(B=0\)