Question

a= 1/1.2 +1/2.3+1/3.4+ ...+ 1/x.x+1

Answer 1

Giải:

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{x\left(x+1\right)}\)

\(\Leftrightarrow A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\)

\(\Leftrightarrow A=\dfrac{1}{1}-\dfrac{1}{x+1}\)

\(\Leftrightarrow A=\dfrac{x}{x+1}\)

Vậy ...

Answer 2

Ta có:

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{x.\left(x+1\right)}\\ A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\\ A=1-\dfrac{1}{x+1}\\ A=\dfrac{x}{x+1}\\ \)

Vậy A=\(\dfrac{x}{x+1}\)