Gọi biểu thức đó là A
\(A=\dfrac{1}{x}.\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+..+\dfrac{1}{49.50}\right)=1\)
\(\Rightarrow A=\dfrac{1}{x}.\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{49.50}\right)=1\)
Ta có công thức : \(\dfrac{a}{b.c}=\dfrac{a}{c-b}.\left(\dfrac{1}{b}-\dfrac{1}{c}\right)\)
Dựa vào công thức trên ta có :
\(\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3}\)
\(\dfrac{1}{3.4}=\dfrac{1}{3}-\dfrac{1}{4}\)
......................
\(\dfrac{1}{49.50}=\dfrac{1}{49}-\dfrac{1}{50}\)
\(\Rightarrow A=\dfrac{1}{x}.\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.....+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\)
\(\Rightarrow A=\dfrac{1}{x}.\left(\dfrac{1}{2}-\dfrac{1}{50}\right)=1\)
\(\Leftrightarrow A=\dfrac{1}{x}.\left(\dfrac{12}{25}\right)=1\)
\(\Rightarrow\) \(\dfrac{1}{x}=A:\dfrac{12}{25}=1:\dfrac{12}{25}=\dfrac{25}{12}\)
Vậy x = 12.
Mink nghĩ vậy, ai thấy đúng thì ủng hộ mink nha !!!
