Tính các số hữu tỉ sau:
A=\(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right).....\left(1-\dfrac{1}{n+1}\right),n\)là số tự nhiên
B=\(\dfrac{1}{2000.1999}-\dfrac{1}{1999.1998}-\dfrac{1}{1998.1997}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
Nhờ các các bn giúp mik!khó quá mik lm hk ra ai giỏi toán giúp mik với!
b,
\(B=\frac{1}{2000.1999}-\frac{1}{1999.1998}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(\Rightarrow B=\frac{1}{1999.2000}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1998.1999}\right)\)
\(\Rightarrow B=\frac{1}{1999.2000}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1998}-\frac{1}{1999}\right)\)
\(\Rightarrow B=\frac{1}{1999.2000}-\left(1-\frac{1}{1999}\right)\)
\(\Rightarrow B=\frac{1}{1999.2000}-\frac{1998}{1999}\)
\(\Rightarrow B=\frac{1}{1999}-\frac{1}{2000}-\frac{1998}{1999}\)
\(\Rightarrow B=\left(\frac{1}{1999}-\frac{1998}{1999}\right)-\frac{1}{2000}\)
\(\Rightarrow B=\frac{-1997}{1999}-\frac{1}{2000}\)
A= (1-\(\dfrac{1}{2}\) ) . ( 1- \(\dfrac{1}{3}\) ) . ( 1 - \(\dfrac{1}{4}\)) ...........( 1- \(\dfrac{1}{n+1}\))
= \(\dfrac{1}{2}\). \(\dfrac{2}{3}\) . \(\dfrac{3}{4}\) ......\(\dfrac{1}{n+1}\)
= 1
Mik nghĩ chắc là thê này đó Khánh Linh
a)
Có:
\(A=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{n+1}\right)\)
\(\Leftrightarrow A=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{n}{n+1}\)
Rút gọn biểu thức A, ta được:
\(A=\dfrac{1}{n-1}\)
Chúc bạn học tốt!