Question

B=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}\)(x thuộc Z)

tính B

các bn giúp mk nhé mk tk

Answer 1

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x.\left(x+1\right)}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}\)

\(=1-\frac{1}{x+1}=\frac{x+1}{x+1}-\frac{1}{x+1}=\frac{x}{x+1}\)

Answer 2

có câu tương tự đó bn^^

Answer 3

có ai lầm đâu mà tương tự

Answer 4

49/50 nhé

tk mk nha! thanks

:))))))))))))))))))

Answer 5

B = \(\frac{1}{1}-\frac{1}{x+1}\)

\(B=\frac{1}{\left(x+1\right)}\)