\(1\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{47.49}=\frac{1}{x}\)

\(1\frac{1}{3}+\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{47.49}\right)=\frac{1}{x}\)

\(\frac{4}{3}+\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{47}-\frac{1}{49}\right)=\frac{1}{x}\)

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

\(\frac{4}{3}+\frac{1}{2}.\frac{46}{147}=\frac{1}{x}\)

\(\frac{4}{3}+\frac{23}{147}=\frac{1}{x}\)

\(\frac{73}{49}=\frac{1}{x}\)

=>\(x=\frac{49.1}{73}=\frac{49}{73}\Rightarrow\)I x I= \(\frac{49}{73}\)

=1-1/3+1/3-1/5+.....+1/47-1/49

=1-1/49

=)x=49

...kcho minh nha

\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{47.49}=\frac{1}{x}\)

\(\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{47.49}\right)=\frac{1}{x}\)

\(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{37}-\frac{1}{49}\right)=\frac{1}{x}\)

\(\frac{1}{2}\left(1-\frac{1}{49}\right)=\frac{1}{x}\)

\(\frac{1}{2}\cdot\frac{48}{49}=\frac{1}{x}\)

\(\frac{1}{x}=\frac{24}{49}\)

=>x=49/24

\(\Leftrightarrow\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{47}-\frac{1}{49}\right)=\frac{1}{x}\)

\(\Leftrightarrow\frac{1}{2}\left(1-\frac{1}{49}\right)=\frac{1}{x}\Rightarrow x=\frac{49}{24}\)

\(\frac{1}{2}.\left(1-\frac{1}{3}\right)+\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+...+\frac{1}{2}.\left(\frac{1}{47}-\frac{1}{49}\right)=\frac{1}{x}\)

\(\frac{1}{2}.\left(1-\frac{1}{49}\right)=\frac{1}{x}\)

\(\frac{24}{49}=\frac{1}{x}\)\(\Rightarrow x=\frac{49}{24}\)

\(K=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{47.49}\)

   \(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{47}-\frac{1}{49}\right)\)

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

   \(=\frac{1}{2}.\frac{48}{49}\)

   \(=\frac{24}{49}\)

\(K\times2=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{47.49}\)

\(K\times2=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{47}-\frac{1}{49}\)

\(K\times2=\frac{48}{49}\)

\(K=\frac{48}{49}\div2=\frac{24}{49}\)

\(K=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+..........+\frac{2}{47.49}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+........+\frac{1}{47}-\frac{1}{49}\right)\)

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

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

\(=\frac{24}{49}\)

\(P=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{47.49}\)

\(P=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...+\frac{1}{47}-\frac{1}{49}\right)\)

\(P=\frac{1}{2}.\left(1-\frac{1}{49}\right)\)

\(P=\frac{1}{2}.\frac{48}{49}\)

\(P=\frac{24}{49}\)

\(P=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{47.49}\)

\(P=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{47.49}\right)\)

\(P=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{47}-\frac{1}{49}\right)\)

\(P=\frac{1}{2}.\left(1-\frac{1}{49}\right)\)

\(P=\frac{1}{2}.\frac{48}{49}\)

\(P=\frac{24}{49}\)

\(K=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{47.49}\)

    \(=2\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{47.49}\right):2\)

     =  \(\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{47}-\frac{1}{49}\right):2\)   

     = \(\left(1-\frac{1}{49}\right):2\)

     \(=\frac{48}{49}:2\) \(\frac{24}{49}\)

\(\frac{48}{49}\)

\(K=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+....+\frac{1}{47}-\frac{1}{49}\right)=\frac{1}{2}.\frac{46}{147}=\frac{23}{147}\)

theo công thức, ta tính đc: 

A = 1- 1/3 + 1/3 - 1/5 + 1/5 -1/7 +..... + 1/49 - 1/51

=> A bằng 1- 1/51 ( các cặp phân số đối nhau thì lược bỏ như - 1/3 và + 1/3 )

1-1/3=2/3 chứ ko phải 1-1/3=1/3 đâu nha bạn

theo bài ra ta có:

A=1-1/3+1/3-1/5+1/5-1/7+......+1/47-1/49+1/49-1/51

A=1-1/51

\(b)\) \(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{97.101}=\frac{2x+4}{101}\)

\(\Leftrightarrow\)\(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{101}=\frac{2x+4}{101}\)

\(\Leftrightarrow\)\(1-\frac{1}{101}=\frac{2x+4}{101}\)

\(\Leftrightarrow\)\(\frac{100}{101}=\frac{2x+4}{101}\)

\(\Leftrightarrow\)\(100=2x+4\)

\(\Leftrightarrow\)\(2x=96\)

\(\Leftrightarrow\)\(48\)

Vậy \(x=48\)

Chúc bạn học tốt ~ 

\(a)\) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{47.49}=\frac{24}{x+1}\)

\(\Leftrightarrow\)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{47.49}=\frac{48}{x+1}\)

\(\Leftrightarrow\)\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{47}-\frac{1}{49}=\frac{48}{x+1}\)

\(\Leftrightarrow\)\(1-\frac{1}{49}=\frac{48}{x+1}\)

\(\Leftrightarrow\)\(\frac{48}{49}=\frac{48}{x+1}\)

\(\Leftrightarrow\)\(49=x+1\)

\(\Leftrightarrow\)\(x=48\)

Vậy \(x=48\)

Chúc bạn học tốt ~ 

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