Bài 28 (Sách bài tập - trang 31)
Gửi bởi: Sách Giáo Khoa Vào 26 tháng 12 2018 lúc 11:16:11
Lý thuyết
Câu hỏi
a) Chứng minh :
\(\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1}{x\left(x+1\right)}\)
b) Đố :
Đố em tính nhẩm được tổng sau :
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{x+5}\)
Hướng dẫn giải
a.\(\dfrac{1}{x}\)-\(\dfrac{1}{x+1}\)=\(\dfrac{x+1}{x\left(x+1\right)}\)-\(\dfrac{x}{x\left(x+1\right)}\)=\(\dfrac{x+1-x}{x\left(x+1\right)}\)=\(\dfrac{1}{x\left(x+1\right)}\)
b. Ta có:
\(\dfrac{1}{x\left(x+1\right)}\)= \(\dfrac{\left(x+1\right)-x}{x\left(x+1\right)}\)=\(\dfrac{x+1}{x\left(x+1\right)}\)-\(\dfrac{x}{x\left(x+1\right)}\)=\(\dfrac{1}{x}\)-\(\dfrac{1}{x+1}\)
Ta lại có:
\(\dfrac{1}{\left(x+1\right)\left(x+2\right)}\)=\(\dfrac{1}{x+1}\)-\(\dfrac{1}{x+2}\);
\(\dfrac{1}{\left(x+2\right)\left(x+3\right)}\)=\(\dfrac{1}{x+2}\)-\(\dfrac{1}{x+3}\);
\(\dfrac{1}{\left(x+3\right)\left(x+4\right)}\)=\(\dfrac{1}{x+3}\)-\(\dfrac{1}{x+4}\);
\(\dfrac{1}{\left(x+4\right)\left(x+5\right)}\)=\(\dfrac{1}{x+4}\)-\(\dfrac{1}{x+5}\);
Do đó:
\(\dfrac{1}{x\left(x+1\right)}\)+\(\dfrac{1}{\left(x+1\right)\left(x+2\right)}\)+\(\dfrac{1}{\left(x+2\right)\left(x+3\right)}\)+\(\dfrac{1}{\left(x+3\right)\left(x+4\right)}\)+\(\dfrac{1}{\left(x+4\right)\left(x+5\right)}\)+\(\dfrac{1}{x+5}\) = \(\dfrac{1}{x}\)-\(\dfrac{1}{x+1}\)+\(\dfrac{1}{x+1}\)-\(\dfrac{1}{x+2}\)+\(\dfrac{1}{x+2}\)-...... -\(\dfrac{1}{x+5}\)+\(\dfrac{1}{x+5}\)=\(\dfrac{1}{x}\)
Vậy tổng trên bằng \(\dfrac{1}{x}\)
Update: 26 tháng 12 2018 lúc 14:57:28
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