\(x^2-x-2001.2002=x^2-x-4006002\)

`x^2-x-2001.2002`

`=x^2-2002x+2001x-2001.2002`

`=x(x-2002)+2001(x-2002)`

`=(x-2002)(x+2001)`.

x² - x - 2001.2002

= (x² - 2002x) + (2001x - 2001.2002)

= x(x - 2002) + 2001(x - 2002)

= (x + 2001)(x- 2002)

x²-x-2001.2002

=x²-x-2001(2001+1)

=x²-x-2001²-2001

=(x²-2001²)-(x+2001)

=(x-2001)(x+2001)-(x+2001)

=(x+2001)(x-2001-1)

=(x+2001)(x-2002)

\(x^2-x-2001.2002\)

= \(x^2+2001x-2002x-2001.2002\)

= \(x\left(x+2001\right)-2002\left(x+2001\right)\)

\(\left(x+2001\right)\left(x-2002\right)\)

a) Ta có: \(x^3+x^2+4\)

\(=x^3+2x^2-x^2+4\)

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

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

b) Ta có: \(9x^2+12x-5\)

\(=9x^2+15x-3x-5\)

\(=3x\left(3x+5\right)-\left(3x+5\right)\)

\(=\left(3x+5\right)\left(3x-1\right)\)

c) Ta có: \(x^4+1997x^2+1996x+1997\)

\(=x^4+x^2+1+1996x^2+1996x+1996\)

\(=\left(x^4+2x^2+1-x^2\right)+1996\left(x^2+x+1\right)\)

\(=\left[\left(x^2+1\right)^2-x^2\right]+1996\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+1\right)+1996\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+1997\right)\)

d) Ta có: \(x^2-x-2001\cdot2002\)

\(=x^2-2002x+2001x-2001\cdot2002\)

\(=x\left(x-2002\right)+2001\left(x-2002\right)\)

\(=\left(x-2002\right)\left(x+2001\right)\)

Hông bít

\(\frac{2001.2002+2003.21+1981}{2002.2003-2001.2002}\)

Đặt x=2001 => x+1=2002 ,x+2=2003,x-20=1981

Thay vào pt \(\Leftrightarrow\frac{x.\left(x+1\right)+21\left(x+2\right)+x-20}{\left(x+1\right)\left(x+2\right)-x\left(x+1\right)}\Leftrightarrow\frac{x^2+x+21x+42+x-20}{x^2+3x+2-x^2-x}=\frac{x^2+23x+22}{2x+2}\)

\(=\frac{x^2+22x+x+22}{2\left(x+1\right)}=\frac{x\left(x+22\right)+\left(x+22\right)}{2\left(x+1\right)}=\frac{\left(x+22\right)\left(x+1\right)}{2\left(x+1\right)}=\frac{x+22}{2}\)

\(=\frac{2001+22}{2}=\frac{2023}{2}=1011,5\)

Chúc bạn học tốt nha . Chọn mình nha Cảm ơn

Cảm ơn bạn bạn học lớp mấy bạn có thể kết bạn với mình không

a) \(\left(x+y+z\right)\left(xy+yz+xz\right)-xyz\)

\(=\left(y+z\right)\left(xy+yz+zx\right)+x^2y+x^2z+xyz-xyz\)

\(=\left(y+z\right)\left(xy+yz+zx\right)+x^2\left(y+z\right)\)

\(=\left(y+z\right)\left(xy+yz+zx+x^2\right)\)

\(=\left(y+z\right)\left[y\left(x+z\right)+x\left(z+x\right)\right]\)

\(=\left(y+z\right)\left(x+z\right)\left(x+y\right)\)

b) \(\left(x^2+y^2+5\right)^2-4x^2y^2-16xy-16\)

\(=\left(x^2+y^2+5\right)^2-\left(4x^2y^2+16xy+16\right)\)

\(=\left(x^2+y^2+5\right)^2-\left(2xy+4\right)^2\)

\(=\left(x^2+y^2+5-2xy-4\right)\left(x^2+y^2+5+2yx+4\right)\)

\(=\left(x^2+y^2+5-2xy-4\right)\left(x^2+y^2+5+2yx+4\right)\)

c)sai đề. 

đặt \(x^2+x+1=t\)

\(\Rightarrow\left(x^2+x+1\right)^2+\left(x^2+x+2\right)-12\)

\(=t^2+t+1-12\)

.........................................

mình sửa đề không biết có đúng hay không nên mình chỉ nêu hướng làm thôi. bạn thông cảm.

d) \(x^2-x-2001.2002\)

\(=x\left(x+2001\right)-2002\left(x+2001\right)\)

\(=\left(x-2002\right)\left(x+2001\right)\)

e)\(x^3-5x^2+8x-4\)

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

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

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

f) \(x^6+x^4-x^2y^2+y^4+y^6\)

\(=\left(x^2\right)^3+\left(y^2\right)^3+\left(x^4-x^2y^2+y^4\right)\)

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

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

(98 . 7676 - 9898. 76) : (2001.2002.....2010)

=(98.101.76-98.101.76) : (2001.2002.....2010)

= 0 : (2001.2002.....2010)

= 0  

(98 . 7676 - 9898. 76) : (2001.2002.....2010)

=(98.101.76-98.101.76):(2001.2002.....2010)

=0:(2001.2002.....2010)

=0

Ta có: \(\left(98\cdot7676-9898\cdot76\right):\left(2001\cdot2002\cdot...\cdot2010\right)\)

\(=98\cdot76\cdot\left(101-101\right)\cdot\left(2001\cdot2002\cdot...\cdot2010\right)\)

=0