\(\text{A = 1-2-3-4+5-6-7-8+9-10-11-12+...........+97-98-99-100}\)

\(\text{A =(1-2-3-4)+(5-6-7-8)+(9-10-11-12)+.............+(97-98-99-100)}\)

\(\text{A =-8+(-16)+(-24)+..................+(-200)}\)

\(\text{A =-8.(1+2+3+......+25)}\)

\(\text{A =-8.[(25-1):1+1.26:2]}\)

\(\text{A =-8.325}\)

\(\text{A =-2600 Vậy A = -2600 }\)

=1+(2-3-4+5)+(6-7-8+9)+...+(98-99-100+101)+102
=1+0+0+0+....+102
=103

k mk nha

ne bang 153 dung khong nhi 

153 ddddo

Đặt \(A=\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{96\cdot98}+\frac{2}{98\cdot100}\)

\(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\)

\(A=\frac{1}{2}-\frac{1}{100}\)

\(A=\frac{50}{100}-\frac{1}{100}=\frac{49}{100}\)

2/2.4 + 2/4.6 + 2/6.8 + ... + 2/98.100

= 1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 + ... + 1/98 - 1/100

= 1/2 + (-1/4 + 1/4) + (-1/6 + 1/6) + ... + (-1/98 + 1/98) - 1/100

= 1/2 - 1/100 = 49/100

Ko viết đàu bài nha!

\(\Rightarrow=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\)

\(\Rightarrow=\frac{1}{2}-\frac{1}{100}\)

\(=\frac{49}{100}\)

2/2.4 + 2/4.6 + 2/6.8 + ..............+ 2/98.100

=2.(1- 1/4 + 1/4 - 1/6 + 1/6 - 1/8+ .........+1/98 - 1/100)

=2.(1-1/100)=2.99/100 = 99/50

phân số nhé

\(\frac{100+98+96+94+...+4+2}{100-98+96-94+...+4-2}\)

\(=\frac{\text{[}\left(100-2\right):1+1\text{]}.102:2}{2+2+2+...+2\left(51s\text{ố}2\right)}\)

\(=\frac{5049}{102}=49\frac{1}{2}\)

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

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

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

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 \(B=0\)

\(S=1+2^2+2^4+...+2^{100}\)

\(2^2S=2^2\cdot\left(1+2^2+2^4+...+2^{100}\right)\)

\(4S=2^2+2^4+...+2^{102}\)

\(4S-S=2^2+2^4+...+2^{102}-1-2^2-...-2^{100}\)

\(3S=2^{102}-1\)

\(S=\dfrac{2^{102}-1}{3}\)