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 \(A=2004\)

5 năm rồi , nếu biết bài này thì chị up hộ em bài giải câu b với =)

thay x=-1,y=1 vào đá thức x⁴.y⁴+x^5.y^5+x^6.y^6+x^7.y^7+x^8.y^8 ta được

=[ (-1)^4.1^4]+[(-1)^5.1^5]+[(-1)^6.1^6]+[(-1)^7.1^7]+[(-1)^8.1^8]

=1+(-1)+1+(-1)+1=1

học tốt nhé!!

Bài 2:

1: \(A=\left(x+2\right)\left(x^2-2x+4\right)+2\left(x+1\right)\left(1-x\right)\)

\(=\left(x+2\right)\left(x^2-x\cdot2+2^2\right)-2\left(x+1\right)\left(x-1\right)\)

\(=x^3+2^3-2\left(x^2-1\right)\)

\(=x^3+8-2x^2+2=x^3-2x^2+10\)

\(B=\left(2x-y\right)^2-2\left(4x^2-y^2\right)+\left(2x+y\right)^2+4\left(y+2\right)\)

\(=\left(2x-y\right)^2-2\cdot\left(2x-y\right)\left(2x+y\right)+\left(2x+y\right)^2+4\left(y+2\right)\)

\(=\left(2x-y-2x-y\right)^2+4\left(y+2\right)\)

\(=\left(-2y\right)^2+4\left(y+2\right)\)

\(=4y^2+4y+8\)

2: Khi x=2 thì \(A=2^3-2\cdot2^2+10=8-8+10=10\)

3: \(B=4y^2+4y+8\)

\(=4y^2+4y+1+7\)

\(=\left(2y+1\right)^2+7>=7>0\forall y\)

=>B luôn dương với mọi y

Bài 1:

5: \(x^2\left(x-y+1\right)+\left(x^2-1\right)\left(x+y\right)\)

\(=x^3-x^2y+x^2+x^3+x^2y-x-y\)

\(=2x^3-x+x^2-y\)

6: \(\left(3x-5\right)\left(2x+11\right)-6\left(x+7\right)^2\)

\(=6x^2+33x-10x-55-6\left(x^2+14x+49\right)\)

\(=6x^2+23x-55-6x^2-84x-294\)

=-61x-349

Ta có:

\(\left(x-1\right)^2+\left(y+2\right)^2=0\)

Do: \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\\\left(y+2\right)^2\ge0\end{matrix}\right.\)

\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0\)

Mặt khác: \(\left(x-1\right)^2+\left(y+2\right)^2=0\)

\(\Rightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

Thay vào B ta có:

\(B=2\cdot1^5-5\cdot\left(-2\right)^3+4=2\cdot1-5\cdot-8+4=2+40+4=46\)

a) M = (x² + 3xy - 3x³) + (2y³ - xy + 3x³)

= x² + 3xy - 3x³ + 2y³ - xy + 3x³

= x² + (3xy - xy) + (-3x³ + 3x³) + 2y³

= x² + 2xy + 2y³

Tại x = 5 và y = 4

M = 5² + 2.5.4 + 2.4³

= 25 + 40 + 2.64

= 65 + 128

= 193

b) N = x²(x + y) - y(x² - y²)

= x³ + x²y - x²y + y³

= x³ + (x²y - x²y) + y³

= x³ + y³

Tại x = -6 và y = 8

N = (-6)³ + 8³

= -216 + 512

= 296

c) P = x² + 1/2 x + 1/16

= (x + 1/2)²

Tại x = 3/4 ta có:

P = (3/4 + 1/2)² = (5/4)² = 25/16

\(A=\dfrac{\dfrac{1}{9}:\dfrac{7}{5}:\dfrac{4}{3}}{\dfrac{1}{81}:\dfrac{49}{25}:\dfrac{16}{9}}=\dfrac{5}{84}:\dfrac{25}{7056}=\dfrac{84}{5}\)