Question

thực hiện phép tính bằng cách hợp lí:

1,A=\(\dfrac{1}{2\cdot3}\)+\(\dfrac{1}{3\cdot4}\)+\(\dfrac{1}{4\cdot5}\)+...+\(\dfrac{1}{99\cdot100}\)

2,B=\(\dfrac{1}{30}\)+\(\dfrac{1}{42}\)+\(\dfrac{1}{56}\)+\(\dfrac{1}{72}\)+\(\dfrac{1}{30}\)+\(\dfrac{1}{110}\)+\(\dfrac{1}{132}\)

Answer 1

1,A=\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+\(\dfrac{1}{4.5}\)+...+\(\dfrac{1}{99.100}\)

1,A= \(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-...-\(\dfrac{1}{99}\)+\(\dfrac{1}{99}\)-\(\dfrac{1}{100}\)

1,A= \(\dfrac{1}{2}\)-\(\dfrac{1}{100}\)

1,A= \(\dfrac{49}{100}\)

(Còn câu 2 mình chưa nghĩ ra. Cho mình hỏi 1 câu, không biết là đề câu 2 có chính xác không?)

Answer 2

2. \(B=\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}\)

⇔ \(B=\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}+\dfrac{1}{10.11}+\dfrac{1}{11.12}\)

⇔ \(B=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{11}-\dfrac{1}{12}\)

⇔ \(B=\dfrac{1}{5}-\dfrac{1}{12}\)

⇔ \(B=\dfrac{7}{60}\)