Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
(2002+1)x14+1988+2001x2002
2002x(1+503+504)
2002x14+14+1988+2001x2002
2002x1008
2002x14+2002+2001x2002
2002x1008
2002x(14+2001+1)
2002x1008
2002x2016
2002x1008
2
1
1988 x 1996 + 1997 x 11 + 1985
= 1988 x1996 + 1996 x 22 + 1985
= 1996 x ( 1988 + 22 ) + 1985
= 1996 x 2000 + 1985
= 3992000 + 1985
= 3993985
\(a,\frac{2001}{2002}.\frac{5}{7}.\frac{2002}{5}.\frac{7}{2001}=\left(\frac{2001}{2002}.\frac{7}{2001}\right).\left(\frac{5}{7}.\frac{2002}{5}\right)\)
\(=\frac{7}{2002}.\frac{2002}{7}=1\)
\(b,\frac{5}{7}.\frac{7}{9}.\frac{9}{11}.\frac{11}{13}=\left(\frac{5}{7}.\frac{7}{9}\right).\left(\frac{9}{11}.\frac{11}{13}\right)=\frac{5}{9}.\frac{9}{13}\)
\(=\frac{5}{13}\)
=\(\frac{2001x2004x1001x2006}{2004x2006x2001x2002}\)=\(\frac{1}{2}\)
\(=\frac{2001.2004.1001.2002}{2004.2006.2001.2002}=\frac{1.1.1001.1}{1.2006.1.1}=\frac{1001}{2006}\)
Ta có :
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\)
\(=\)\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}\)
\(=\)\(1-\frac{1}{2014}\)
\(=\)\(\frac{2014}{2014}-\frac{1}{2014}\)
\(=\)\(\frac{2013}{2014}\)
Vậy \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}=\frac{2013}{2014}\)
Dấu \(.\) là dấu nhân nhé
Chúc bạn học tốt ~
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{2013\times2014}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}\)
\(=1-\frac{1}{2014}\)
\(=\frac{2013}{2014}\)
CHÚC BN HỌC TỐT!!!!!
\(\frac{2003x14+1988+2001x2002}{2002+2002x503+504x2002}\)= \(\left(\frac{2003x14+1988}{2002+2002x503}\right)+\frac{2001x2002}{504x2002}\) = \(\frac{5}{168}+\frac{667}{168}\)= \(\frac{672}{168}=4\)