A = \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}}{\frac{1}{49}+\frac{2}{48}+\frac{3}{47}+...+\frac{48}{2}+\frac{49}{1}}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}}{\left(\frac{1}{49}+1\right)+\left(\frac{2}{48}+1\right)+\left(\frac{3}{47}+1\right)+...+\left(\frac{48}{2}+1\right)+\frac{50}{50}}\)

A = \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}}{\frac{50}{49}+\frac{50}{48}+\frac{50}{47}+...+\frac{50}{2}+\frac{50}{50}}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}}{\left(\frac{1}{49}+\frac{1}{48}+\frac{50}{47}+...+\frac{1}{2}+\frac{1}{50}\right).50}=\frac{1}{50}\)

\(A=\frac{T}{M}\)

\(M=\frac{1}{49}+1+\frac{2}{48}+1+\frac{3}{47}+1+.........+\frac{48}{2}+1+1\)

\(=\frac{50}{49}+\frac{50}{48}+\frac{50}{47}+.........+\frac{50}{2}+1\)

\(=50.\left(\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+......+\frac{1}{2}+\frac{1}{50}\right)=50.T\)

\(A=\frac{T}{50T}=\frac{1}{50}\)

     \(\frac{49}{1}+\frac{48}{2}+\frac{47}{3}+...+\frac{2}{48}+\frac{1}{49}\)

\(=1+1+...+1+\frac{48}{2}+\frac{47}{3}+...+\frac{2}{48}+\frac{1}{49}\)(có 49 số 1)

\(=\left(1+\frac{48}{2}\right)+\left(1+\frac{47}{3}\right)+...+\left(1+\frac{2}{48}\right)+\left(1+\frac{1}{49}\right)+1\)

\(=\frac{50}{2}+\frac{50}{3}+...+\frac{50}{48}+\frac{50}{49}+\frac{50}{50}\)

\(=50\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{50}\right)\)

Chúc bạn học tốt.

p=\(\frac{1}{49}+\frac{2}{48}+\frac{3}{47}+...+\frac{48}{2}+49\)

=\(\left(\frac{1}{49}+1\right)+\left(\frac{2}{48}+1\right)+\left(1+\frac{3}{47}\right)+...+\left(1+\frac{48}{2}\right)+\frac{50}{50}\)

=\(\frac{50}{50}+\frac{50}{49}+\frac{50}{48}+...+\frac{50}{2}\)

=\(50\left(\frac{1}{50}+\frac{1}{49}+\frac{1}{48}+...+\frac{1}{2}\right)\)

p=50*S

\(\frac{S}{\text{p}}=\frac{1}{50}\)

s=1,p=50

Ta có: 

1/49 + 1 = 50/49 

2/48 + 1 = 50/48 

3/47 + 1 = 50/47 
. 
. 
. 
47/3 + 1 = 50/3 

48/2 + 1 = 50/2 

0 + 1 = 50/50 

Cộng vế theo vế dãy đẳng thức trên ta được: 

1/49 + 2/48 +........+ 47/3 + 48/2 + 49 = 50/2 + 50/3 + 50/4 +........+ 50/49 + 50/50 

⇒ 1/49 + 2/48 +........+ 47/3 + 48/2 + 49 = 50 x (1/2 + 1/3 + 1/4 +........+ 1/49 + 1/50) 

⇒ B = 50A 

⇒ A/B = 1/50 

 

Ta có: 

1/49 + 1 = 50/49 

2/48 + 1 = 50/48 

3/47 + 1 = 50/47 
. 
. 
. 
47/3 + 1 = 50/3 

48/2 + 1 = 50/2 

0 + 1 = 50/50 

Cộng vế theo vế dãy đẳng thức trên ta được: 

1/49 + 2/48 +........+ 47/3 + 48/2 + 49 = 50/2 + 50/3 + 50/4 +........+ 50/49 + 50/50 

⇒ 1/49 + 2/48 +........+ 47/3 + 48/2 + 49 = 50 x (1/2 + 1/3 + 1/4 +........+ 1/49 + 1/50) 

⇒ B = 50A 

⇒ A/B = 1/50 

Ta có:\(P=\frac{1}{49}+\frac{2}{48}+\frac{3}{47}+....+\frac{48}{2}+\frac{49}{1}+50-50\)

               \(=\left(1+\frac{1}{49}\right)+\left(1+\frac{2}{48}\right)+\left(1+\frac{3}{47}\right)+...+\left(1+\frac{48}{2}\right)+\left(1+\frac{49}{2}\right)-50\)

              \(=\frac{50}{49}+\frac{50}{48}+\frac{50}{47}+....+\frac{50}{2}+\frac{50}{1}-50\)

              \(=50\left(\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+....+\frac{1}{2}\right)+50-50\)

              \(=50\left(\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+....+\frac{1}{2}\right)\)

mà  \(S=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{49}\)

\(=>\frac{S}{P}=\frac{1}{50}\)

Vậy \(\frac{S}{P}=\frac{1}{50}\)              

\(P=\frac{99}{50}-\frac{97}{49}+...+\frac{7}{4}-\frac{5}{3}+\frac{3}{2}-1\)

\(=2.\left(\frac{99}{100}-\frac{97}{98}+...+\frac{7}{8}-\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)\)

\(=2\left[\left(1-\frac{1}{100}\right)-\left(1-\frac{1}{98}\right)+...+\left(1-\frac{1}{8}\right)-\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{4}\right)-\left(1-\frac{1}{2}\right)\right]\)

Ta có: P = \(\frac{1}{49}+\frac{2}{48}+\frac{3}{47}+...+\frac{49}{1}\)

\(=\frac{49}{1}+\frac{48}{2}+\frac{47}{3}+...+\frac{1}{49}\)

\(=\frac{50-1}{1}+\frac{50-2}{2}+\frac{50-3}{3}+...+\frac{50-49}{49}\)

\(=\frac{50}{1}-\frac{1}{1}+\frac{50}{2}-\frac{2}{2}+\frac{50}{3}-\frac{3}{3}+...+\frac{50}{49}-\frac{49}{49}\)

\(=\left(\frac{50}{1}+\frac{50}{2}+\frac{50}{3}+...+\frac{50}{49}\right)-\left(\frac{1}{1}+\frac{2}{2}+\frac{3}{3}+...+\frac{49}{49}\right)\)

\(=50+50\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}\right)-49\)

\(=50\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}\right)+1\)

\(=50\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}\right)+\frac{50}{50}\)

\(=50\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)\)

\(\Rightarrow\frac{S}{P}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{50}}{50\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)}=\frac{1}{50}\)