a: \(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)

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

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

=>\(3A=2^{101}-2\)

=>\(A=\dfrac{2^{101}-2}{3}\)

b: Sửa đề: \(A=\dfrac{2\cdot8^4\cdot27^2+4\cdot6^9}{2^7\cdot6^7+2^7\cdot40\cdot9^4}\)

\(A=\dfrac{2\cdot2^{12}\cdot3^6+2^2\cdot2^9\cdot3^9}{2^7\cdot2^7\cdot3^7+2^7\cdot2^3\cdot5\cdot3^8}\)

\(=\dfrac{2^{11}\cdot3^6\left(2^3+3^3\right)}{2^{10}\cdot3^7\left(2^4+5\cdot3\right)}\)

\(=\dfrac{2}{3}\cdot\dfrac{4+27}{16+15}=\dfrac{2}{3}\)

c: \(B=\dfrac{4^5\cdot9^4-2\cdot6^4}{2^{10}\cdot3^8+6^8\cdot20}\)

\(=\dfrac{2^{10}\cdot3^8-2\cdot2^4\cdot3^4}{2^{10}\cdot3^8+2^8\cdot2^2\cdot5\cdot3^8}\)

\(=\dfrac{2^5\cdot3^4\left(2^5\cdot3^4-1\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{1}{2^5\cdot3^4}\cdot\dfrac{32\cdot81-1}{6}\)

\(=\dfrac{2591}{2^6\cdot3^5}\)

help

\(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{97}+\frac{1}{99}\)

\(=2-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}\)

\(=2-\frac{1}{99}\)

\(=\frac{197}{99}\)

1+1/3+1/5...+1/97+1/99=(1+1/99) + (1/3+1/97) + (1/5+1/95)....+(1/49+1/51) 
= 100/1.99 + 100/3.97 + 100/5.95 +.....=100.(1/1.99 + 1/3.97 + 1/5.95 +.....) 
Mau so: 
1/1.99 + 1/3.97 +1/5.95....+1/95.5+ 1/97.3 +1/99.1=2/1.99 +2/3.97 +2/5.95+..... 
=2.(1/1.99 + 1/3.97 + 1/5.95 +.....) 
=>A=(100.(1/1.99 + 1/3.97 + 1/5.95 +.....)) : (2.(1/1.99 + 1/3.97 + 1/5.95 +.....))=50

chuẩn luôn , tích nha

Thanks nhìu  ^_^

đúng ko hoài mình cho bạn nhé

ai tích mình mình tích lại

Ta thấy: 
1/1 + 1/99 = (99+1)/(1.99)=100/(1.99) 
1/3 + 1/97 = (97+3)/(3.97)=100/(3.97) 
1/5 + 1/95 = (95+5)/(5.95)=100/(3.97) 
… 
1/97 + 1/3 = (3+97)/(97.3)=100/(97.3) 
1/99 + 1/1 = (1+99)/(99.1)=100/(99.1) 
=> 
1/(1.99)=(1/1+1/99)/100 
1/(3.97)=(1/3+1/97)/100 
… 
1/(99.1)=(1/99+1/1)/100 
------------------------------ cộng 2 vế của các đẳng thức trên. Ta được đẳng thức: 
1/(1.99) + 1/(3.97)+ 1/(5.95) +...+ 1/(97.3) + 1/(99.1 ) 
=[(1/1+1/99)+(1/3+1/99)+…+(1/99+1/1)]/1... 
=2(1+1/3+1/5+1/7…+1/99]/100 
=(1+1/3+1/5+1/7…+1/99]/50 
Vậy: 
A=(1+1/3+1/5+1/7+...+1/97+1/99) / [ 1/(1.99) + 1/(3.97)+ 1/(5.95) +...+ 1/(97.3) + 1/(99.1 ) ] 
A=(1+1/3+1/5+1/7+...+1/97+1/99)/[(1+1/3... 
A=50. 

bạn nói là gnoo Mình Hoàng đúng ko

50 la dung do

\(A=2^{100}-\left(2^{99}+2^{98}+...+2+1\right)\)

Đặt \(B=2^{99}+2^{98}+...+2+1\)

\(\Rightarrow2B=2^{100}+2^{99}+...+2^2+2\)

\(\Rightarrow2B-B=2^{100}-1\Leftrightarrow B=2^{100}-1\)

\(\Rightarrow A=2^{100}-\left(2^{100}-1\right)=1\)

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a) \(A=1+2+2^2+...+2^{50}\)

\(\Rightarrow2A=2+2^2+...+2^{51}\)

\(\Rightarrow A=2A-A=2+2^2+...+2^{51}-1-2-2^2-...-2^{50}=2^{51}-1\)

b) \(B=1+3+3^2+...+3^{100}\)

\(\Rightarrow3B=3+3^2+...+3^{101}\)

\(\Rightarrow2B=3B-B=3+3^2+...+3^{101}-1-3-3^2-...-3^{100}=3^{101}-1\)

\(\Rightarrow B=\dfrac{3^{101}-1}{2}\)

c) \(C=5+5^2+...+5^{30}\)

\(\Rightarrow5C=5^2+5^3+...+5^{31}\)

\(\Rightarrow4C=5C-C=5^2+5^3+...+5^{31}-5-5^2-...-5^{30}=5^{31}-5\)

\(\Rightarrow C=\dfrac{5^{31}-5}{4}\)

d) \(D=2^{100}-2^{99}+2^{98}-...+2^2-2\)

\(\Rightarrow2D=2^{101}-2^{100}+2^{99}-...+2^3-2^2\)

\(\Rightarrow3D=2D+D=2^{101}-2^{100}+2^{99}-...+2^3-2^2+2^{100}-2^{99}+...+2^2-2=2^{101}-2\)

\(\Rightarrow D=\dfrac{2^{101}-2}{3}\)