16700/27

mình chịu

= 200400/324 ( chưa rút gọn)

\(\frac{2004.37+2004+2.2004+2004.59+2004}{324.321-201.324-101.324-18.324}\)

\(=\frac{2004.\left(37+1+2+59+1\right)}{324.\left(321-201-101-18\right)}\)

\(=\frac{2004.100}{324.1}\)

\(=\frac{200400}{324}\)

\(=\frac{16700}{27}\)

dấu bài chưa học

Thay \(x=2003\) vào A ta có:\(A=2003^{17}-2004.2003^{16}+2004.2003^{15}-2004.2003^{14}+...+2004.\left(2003-1\right)\)

\(=2003^{17}-\left(2003+1\right).2003^{16}+\left(2003+1\right).2003^{15}-\left(2003+1\right).2003^{14}+...+\left(2003+1\right).\left(2003-1\right)\)

\(=2003^{17}-2003^{17}+2003^{16}-2003^{16}+2003^{15}-2003^{15}+2003^{14}-2003^{14}+...+\left(2003+1\right).\left(2003-1\right)\)

\(=2004.2002=4012008\)

Dãy trên có 2004x

2004x+(1+2+3+4+...+2003)=2004

2004x+[(2003+1)x{(2003-1):1+1]:2}=2004

2004x+2007006=2004

2004x=2004-2007006

2004x=-2005002

x=-2005002:2004

x=-2001/2

Chúc em học tốt^^

f(x)=\(x^{17}-2004.x^{16}+2004.x^{15}-2004.x^{2014}+...+2004.x-1\)

= \(x^{17}-\left(2003+1\right)x^{16}+\left(2003+1\right)x^{15}-\left(2003+1\right)^{14}+...+\left(2003+1\right)-1\)

Thay x = 2003

=> f(x)= \(x^{17}-\left(x+1\right)x^{16}+\left(x+1\right)x^{15}-\left(x+1\right)x^{14}+...+\left(x+1\right)x-1\)

=\(x^{17}-x^{17}-x^{16}+x^{16}+x^{15}-x^{15}-x^{14}+...+x^2+x-1\)

= \(x-1\)

= 2003 -1

=2002