\(\text{Ta có​: }\)​:     \(\frac{2006}{2008}\text{ x }\frac{2001}{2004}\text{ x }\frac{2008}{2002}\text{ x }\frac{2004}{2006}\text{ x }\frac{1001}{2001}\)

\(=\frac{2006\text{ x }\text{ }2001\text{ x }2008\text{ x }2004\text{ x }1001}{2008\text{ x }2004\text{ x }2002\text{ x }2006\text{ x }2001}\text{ }\)

Rút gọn các số ( 2006 ; 2001 ; 2008 ; 2004) ở cả tử và mẫu ta có:

\(=\frac{1001}{2002}=\frac{1}{2}\)

\(\frac{2006}{2008}\times\frac{2001}{2004}\times\frac{2008}{2002}\times\frac{1001}{2001}\)

\(=\frac{2006\times2001\times2008\times1001}{2008\times2004\times2002\times2001}\)

\(=\frac{1001}{2002}=\frac{1}{2}\)

\(\frac{2006}{2008}\times\frac{2001}{2004}\times\frac{2008}{2002}\times\frac{2004}{2006}\times\frac{1001}{2001}=\frac{2006.2001.2008.2004.1001}{2008.2004.2002.2006.2001}\)

\(=\frac{\left(2001.2004.2006.2008.\right).1001}{\left(2001.2004.2006.2008\right).2002}=\frac{1001}{2002}=\frac{1001.1}{1001.2}=\frac{1}{2}\)

Sao nó cứ khó see kiểu j đấy cậu ak!_-_

Câu trả lời của bạn Forever Alone khó hiểu quá !!!!!!!!!!!

=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!!!

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.

=\(\frac{2001x2004x1001x2006}{2004x2006x2001x2002}\)=\(\frac{1}{2}\)

\(=\frac{2001.2004.1001.2002}{2004.2006.2001.2002}=\frac{1.1.1001.1}{1.2006.1.1}=\frac{1001}{2006}\)

mình ghi nhầm đề

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

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=\(\frac{1}{9}\)

c=\(\frac{3069}{500}\)

= 98/45-31/15

=1/9

x=-2007