Question

 Tớ đố bạn nào làm được nhé :

\(\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\right)x=1\)

Answer 1

\(\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\right)x=1\)

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

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

\(\frac{12}{25}.x=1\)

\(\Rightarrow x=1:\frac{12}{25}\)

\(\Rightarrow x=\frac{25}{12}\)

Answer 2

( \(\frac{1}{2.3}+...+\)\(\frac{1}{49.50}\)) x = 1

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

( \(1-\frac{1}{50}\)) x = 1

    \(\frac{49}{50}\). x = 1 

                    x = 1 : \(\frac{49}{50}\)

                    x = \(\frac{50}{49}\)

           Vậy x = \(\frac{50}{49}\)

Answer 3

\(\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\cdot\cdot\cdot+\frac{1}{49\cdot50}\right)x=1\)

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

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

\(\frac{12}{25}x=1\)

\(x=\frac{25}{12}\)

Answer 4

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

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

\(\left[\frac{1}{2}+0+0+0+...+0-\frac{1}{50}\right]x=1\)

\(\left[\frac{1}{2}-\frac{1}{50}\right]x=1\)

\(\left(\frac{25}{50}-\frac{1}{50}\right)x=1\)

\(=\frac{12}{25}x=1\)

\(x=1:\frac{12}{25}=\frac{25}{12}=2\frac{1}{12}\)

Vậy \(x=2\frac{1}{12}\) 

Answer 5

Dễ ợt

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

\(\Rightarrow\frac{24}{50}x=1\Rightarrow x=1\div\frac{12}{25}=\frac{25}{12}\)

Answer 6

What??????????????????????????????????????????????????

Answer 7

dễ vậy cũng đòi đố

Answer 8

Oh ! ĐM