\(\frac{2006}{2008}\)X\(\frac{2001}{2004}\) X \(\frac{2008}{2002}\) X \(\frac{2004}{2006}\) X \(\frac{1001}{2001}\)
Tính dùm mình nha
tính nhanh
c)\(\frac{2006}{2008}x\frac{2001}{2004}x\frac{2008}{2002}x\frac{2004}{2006}x\frac{1001}{2001}\)
. là x á nha
=\(\frac{2006}{2008}.\frac{2001}{2004}.\frac{2008}{2002}.\frac{2004}{2006}.\frac{1001}{2001}\)
=\(\frac{2006.2001.2008.2004.1001}{2008.2004.2002.2006.2001}\)
=\(\frac{1001}{2002}\)
= \(\frac{2006\cdot2001\cdot2008\cdot2004\cdot1001}{2008\cdot2004\cdot2002\cdot2006\cdot2001}\)
= \(\frac{1\cdot1\cdot1\cdot1\cdot1001}{1\cdot1\cdot2002\cdot1\cdot1}\)
= \(\frac{1}{2}\)
\(\frac{2006}{2008}.\frac{2001}{2004}.\frac{2008}{2002}.\frac{2004}{2006}\frac{1001}{2001}\)
\(=\frac{2006.2001.2008.2004.1001}{2008.2004.2002.2006.2001}\)
\(=\frac{1001}{2002}\)
\(=\frac{1}{2}\)
tinh nhanh
2006/2008 x 2001/2004 x 2008/2002 x 2004/2006 x 1001/2001
\(\frac{2006}{2008}\times\frac{2001}{2004}\times\frac{2008}{2002}\times\frac{2004}{2006}\times\frac{1001}{2001}=\frac{2006.2001.2008.2004.1001}{2008.2004.2002.2006.2001}\)
\(=\frac{\left(2001.2004.2006.2008.\right).1001}{\left(2001.2004.2006.2008\right).2002}=\frac{1001}{2002}=\frac{1001.1}{1001.2}=\frac{1}{2}\)
Sao nó cứ khó see kiểu j đấy cậu ak!_-_
Câu trả lời của bạn Forever Alone khó hiểu quá !!!!!!!!!!!
So sành \(\frac{2002}{2001}+\frac{2003}{2002}+\frac{2004}{2003}+\frac{2005}{2004}+\frac{2006}{2005}+\frac{2007}{2006}+\frac{2008}{2007}+\frac{2009}{2008}\)với 8
=1+1/2001+1+1/2002+1+1/2003+...+1+1/2008=8+1/2001+1/2002+1/2003+...+1/2008>8
\(\frac{2002}{2001}+\frac{2003}{2002}+\frac{2004}{2003}+\frac{2005}{2004}+\frac{2006}{2005}+\frac{2007}{2006}+\frac{2008}{2007}+\frac{2009}{2008}>8\)
Ta có:
2002/2001=1+1/2001
2003/2002=1+1/2002
2004/2003= 1+ 1/2003
2005/2004= 1+ 1/2004
2006/2005=1+ 1/2005
2007/2006= 1+ 1/2006
2008/2007=1 + 1/2007.
2009/2008=1+ 1/2008.
=> 2002/2001+2003/2002+2004?2003+2005/2004+2006/2005+ 2007/2006+ 2008/2007+ 2009/2008= 1+1+1+1+1+1+1+1+1/2001+1/2002+1/2003+1/2004+1/2005+1/2006+1/2007+1/2008>8.
Nhớ k đúng cho mình nha!! Thanks!!!
bn nào giúp mik giải phép tính này vs cảm ơn trước nha nhớ kb vs mik đó!!!!!
\(\frac{2001}{2004}\)x\(\frac{2004}{2006}\)x\(\frac{1001}{2001}\)x\(\frac{2006}{2002}\)=?
=\(\frac{2001x2004x1001x2006}{2004x2006x2001x2002}\)=\(\frac{1}{2}\)
\(=\frac{2001.2004.1001.2002}{2004.2006.2001.2002}=\frac{1.1.1001.1}{1.2006.1.1}=\frac{1001}{2006}\)
\(B=\frac{2009-\frac{2009}{2001}-\frac{2009}{2002}-\frac{2009}{2003}-\frac{2009}{2004}}{2010-\frac{2010}{2001}-\frac{2010}{2002}-\frac{2010}{2003}-\frac{2010}{2004}}:\frac{2009-\frac{2009}{2005}-\frac{2009}{2006}-\frac{2009}{2007}-\frac{2009}{2008}}{2010-\frac{2010}{2005}-\frac{2010}{2006}-\frac{2010}{2007}-\frac{2010}{2008}}\)
minh lam duoc roi . cach viet phan so ban bam vao o mau vang o cuoi trang .cu di con chuot xuong cuoi trang thi thay 1 o vang , vao xem huong dan la biet ngay ma.
Cho A = \(\frac{2000}{2001}+\frac{2001}{2002}+\frac{2002}{2003}+\frac{2003}{2004}+\frac{2005}{2006}+\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}+\frac{2011}{2012}+\frac{2012}{2013}+\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}\)
Hãy so sánh tổng các phân số trong A và so sánh với 15.
mỗi số hạng trong biểu thức A đều nhỏ hơn 1 mà có 15 số nên tổng A sẽ nhỏ hơn 15
ta thay tong tren <1+1+1+1+1+1+1+1+1+1+1+1+1+1+1
hay tong tren be hon 15
Giải phương trình sau :
\(\frac{x^2-2008}{2007}+\:\frac{x^2-2007}{2006}+\frac{x^2-2006}{2005}=\:\frac{x^2-\:2005}{2004}+\:\frac{x^2-2004}{2003}+\:\frac{x^2-2003}{2002}\)
Ta có : \(\frac{x^2-2008}{2007}+\frac{x^2-2007}{2006}+\frac{x^2-2006}{2005}=\frac{x^2-2005}{2004}+\frac{x^2-2004}{2003}+\frac{x^2-2003}{2002}\)
=> \(\frac{x^2-2008}{2007}+1+\frac{x^2-2007}{2006}+1+\frac{x^2-2006}{2005}+1=\frac{x^2-2005}{2004}+1+\frac{x^2-2004}{2003}+1+\frac{x^2-2003}{2002}+1\)
=> \(\frac{x^2-2008}{2007}+\frac{2007}{2007}+\frac{x^2-2007}{2006}+\frac{2006}{2006}+\frac{x^2-2006}{2005}+\frac{2005}{2005}=\frac{x^2-2005}{2004}+\frac{2004}{2004}+\frac{x^2-2004}{2003}+\frac{2003}{2003}+\frac{x^2-2003}{2002}+\frac{2002}{2002}\)
=> \(\frac{x^2-1}{2007}+\frac{x^2-1}{2006}+\frac{x^2-1}{2005}=\frac{x^2-1}{2004}+\frac{x^2-1}{2003}+\frac{x^2-1}{2002}\)
=> \(\frac{x^2-1}{2007}+\frac{x^2-1}{2006}+\frac{x^2-1}{2005}-\frac{x^2-1}{2004}-\frac{x^2-1}{2003}-\frac{x^2-1}{2002}=0\)
=> \(\left(x^2-1\right)\left(\frac{1}{2007}+\frac{1}{2006}+\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}-\frac{1}{2002}\right)=0\)
=> \(x^2-1=0\)
=> \(x^2=1\)
=> \(x=\pm1\)
Vậy phương trình có 2 nghiệm là x = 1, x = -1 .
a 989898/454545 - 31313131/15151515
b 10101x < 5/10101 + 5/20202 + 5/30303 + 5/40404
c 1/1000 + 13/1000 + 25/1000 + 37/1000 + 49/1000 + ..........+ 87/1000 + 99/1000.
d 2006/2008 x 2001/2004 x 2008/2002 x 2004/2006 x 1001/2001
a=\(\frac{1}{9}\)
c=\(\frac{3069}{500}\)
\(\frac{x+6}{2001}+\frac{x+5}{2002}+\frac{x+4}{2003}=\frac{x+3}{2004}+\frac{x+2}{2005}\)+\(\frac{x+1}{2006}\)
Tìm x