Question

Tính giá trị biểu thức:

A= (a+1)(a+2)(a+3)....(a+2003)(a+2004)/(b+5)(b+6)(b+7)....(b+2006)(b+2007) tại a= 0, b= -4

 

B= 1/(x−5)(y+7)+ 1/(x−4)(y+8)+....+ 1/(x−1)(y+11) tại x= 6, y= -5

Answer 1

Ta có : 

\(A=\frac{\left(a+1\right)\left(a+2\right)\left(a+3\right).....\left(a+2003\right)\left(a+2004\right)}{\left(b+5\right)\left(b+6\right)\left(b+7\right).....\left(b+2006\right)\left(b+2007\right)}\)

\(\Leftrightarrow\)\(A=\frac{\left(0+1\right)\left(0+2\right)\left(0+3\right).....\left(0+2003\right)\left(0+2004\right)}{\left(-4+5\right)\left(-4+6\right)\left(-4+7\right).....\left(-4+2006\right)\left(-4+2007\right)}\)

\(\Leftrightarrow\)\(A=\frac{1.2.3.....2003.2004}{1.2.3.....2002.2003}\)

\(\Leftrightarrow\)\(A=\frac{1.2.3.....2003}{1.2.3.....2003}.2004\)

\(\Leftrightarrow\)\(A=2004\)
 

Vậy \(A=2004\)