\(\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right)x=1\)
\(\Rightarrow\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)x=1\)
\(\Rightarrow\left(\dfrac{1}{2}-\dfrac{1}{50}\right)x=1\)
\(\Rightarrow\dfrac{12}{25}x=1\)
\(\Rightarrow x=\dfrac{25}{12}\)
Vậy \(x=\dfrac{25}{12}\)
\(\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right).x=1\)
Ta có: \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)
\(=\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{50-49}{49.50}\)
\(=\dfrac{3}{2.3}-\dfrac{2}{2.3}+\dfrac{4}{3.4}-\dfrac{3}{3.4}+...+\dfrac{50}{49.50}-\dfrac{49}{49.50}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=\dfrac{1}{2}-\dfrac{1}{50}=\dfrac{12}{25}\)
\(\Rightarrow\dfrac{12}{25}.x=1\Rightarrow x=1:\dfrac{12}{25}=\dfrac{25}{12}=2\dfrac{1}{12}\)
Vậy \(x=\dfrac{25}{12}\) hay \(x=2\dfrac{1}{12}\)
\(\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right)x=1\)
\(\Rightarrow\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)x=1\)
\(\Rightarrow\left(\dfrac{1}{2}-\dfrac{1}{50}\right)x=1\)
\(\Rightarrow\dfrac{12}{25}x=1\)
\(\Rightarrow x=\dfrac{25}{12}\)
\(\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1:x\)
Hay \(1:x=\dfrac{1}{2}-\dfrac{1}{50}\)
\(1:x=\dfrac{23}{50}\)
\(x=\dfrac{23}{50}.1\)
\(x=\dfrac{23}{50}\)
Vậy \(x=\dfrac{23}{50}\)
tick cho mk nha!
\(\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right)x=1\\ \Rightarrow\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{49}-\dfrac{1}{50}\right)x=1\\ \Rightarrow\left(\dfrac{1}{2}-\dfrac{1}{50}\right)x=1\\ \Rightarrow\dfrac{12}{25}x=1\\ \Rightarrow x=\dfrac{25}{12}\)
\(\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right).x=1\)
\(\left(\dfrac{1}{2}-\dfrac{1}{50}\right).x=1\)
\(\dfrac{12}{25}.x=1\)
x = 1: \(\dfrac{12}{25}\)
x = \(\dfrac{25}{12}\)
Vậy x = \(\dfrac{25}{12}\)
(\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)).x=1
=(\(\dfrac{1}{2}+\dfrac{1}{50}\)).x=1
=\(\dfrac{25}{12}\).X=1
\(\Rightarrow\)x=1:\(\dfrac{25}{12}\)
\(\Rightarrow\)x=\(\dfrac{12}{25}\)
chúc bn hc giỏi