=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

\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{47.49}=\frac{1}{x}\\ \frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{47.49}\right)=\frac{1}{x}\\ \frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{45}-\frac{1}{47}+\frac{1}{47}-\frac{1}{49}\right)=\frac{1}{x}\)

\(\frac{1}{2}.\left(1-\frac{1}{49}\right)=\frac{1}{x}\\ \frac{1}{2}-\frac{1}{98}=\frac{1}{x}\\ \frac{49-1}{98}=\frac{1}{x}\\ \frac{24}{49}=\frac{1}{x}\\ \Rightarrow24x=49\\ x=\frac{49}{24}\\ x=2\frac{1}{24}\)

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

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

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

   \(=\frac{1}{2}.\frac{48}{49}\)

   \(=\frac{24}{49}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Ta có: \(\left(\frac{10}{1.2}+\frac{10}{2.3}+...+\frac{10}{49.50}\right)+2x=\frac{4}{1.3}+\frac{4}{3.5}+...+\frac{4}{47.49}-7x\) (1)

Xét vế trái ta có:

\(\left(\frac{10}{1.2}+\frac{10}{2.3}+...+\frac{10}{49.50}\right)+2x\)

\(=10.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\right)\)

\(=10.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)+2x\)

\(=10.\left(1-\frac{1}{50}\right)+2x\)

\(=10.\frac{49}{50}+2x\)

\(=\frac{49}{5}+2x\) (2)

Xét vế phải ta có:

\(\frac{4}{1.3}+\frac{4}{3.5}+...+\frac{4}{47.49}-7x\)

\(=2.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{47.49}\right)-7x\)

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

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

\(=2.\frac{48}{49}-7x\)

\(=\frac{96}{49}-7x\) (3)

Từ (1), (2) và (3) => \(\frac{49}{5}+2x=\frac{96}{49}-7x\)

\(\Rightarrow2x+7x=\frac{96}{49}-\frac{49}{5}\)

\(\Rightarrow9x=\frac{480}{245}-\frac{2401}{245}\)

\(\Rightarrow9x=-\frac{1921}{245}\)

\(\Rightarrow x=-\frac{1921}{245}:9=-\frac{1921}{2205}\)

Vậy \(x=-\frac{1921}{2205}\)

Chúc bạn học tốt!

Ta có:\(\left(10-\frac{10}{2}+\frac{10}{2}-\frac{10}{3}+...+\frac{10}{49}-\frac{10}{50}\right)+2x=\left(2-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+...+\frac{2}{47}-\frac{2}{49}\right)-7x\)

          \(\left(10-\frac{10}{50}\right)+2x=\left(2-\frac{2}{49}\right)-7x\)

           \(\frac{49}{5}+2x=\frac{96}{49}-7x\)

            \(7x+2x=\frac{96}{49}-\frac{49}{5}\)

            \(9x=-\frac{1921}{245}\)

            \(x=-\frac{1921}{245}:9\)

            \(x=-\frac{1921}{2205}\)

Vậy \(x=-\frac{1921}{2205}\)