Bài 1:
a) \(A=x\left(x-1\right)-y\left(x-1\right)\)
\(=\left(x-1\right)\left(x-y\right)\)
Thay x = 2001; y = 1999 vào A, ta được:
\(A=\left(2001-1\right)\left(2001-1999\right)\)
\(=2000.2\)
\(=4000\)
b) \(B=-45x^4y^3z^2:15xy^2z^2\)
\(=\left(-45:15\right).\left(x^4:x\right).\left(y^3:y^2\right).\left(z^2:z^2\right)\)
\(=-3x^3y\)
Thay x = 2; y = 10 vào B, ta được:
\(B=-3.2^3.10=-240\)
\(A=x\left(x-1\right)-y\left(x-1\right)\)
\(=\left(x-y\right)\left(x-1\right)\)
Thay x=2001, y=1999, ta có:
\(A=\left(2001-1999\right)\left(2001-1\right)\)
\(=2.2000=4000\)
`A = (x-1)(x-y)`
`= (2001-1)(2001-1999)`
`= 2 . 2000`
`= 4000`
`B = -3x^3y = -3 . 8 . 10 = -240`
Bài 2
a) \(3x^2-6xy+3y^2\)
\(=3\left(x^2-2xy+y^2\right)\)
\(=3\left(x-y\right)^2\)
b) \(x^2+4y-y^2-4x\)
\(=\left(x^2-y^2\right)-\left(4x-4y\right)\)
\(=\left(x-y\right)\left(x+y\right)-4\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-4\right)\)
c) \(-9x^2+6x+y^2-1\)
\(=-\left(9x^2-6x+1-y^2\right)\)
\(=-\left[\left(3x-1\right)^2-y^2\right]\)
\(=-\left(3x-y-1\right)\left(3x+y-1\right)\)
giúp em với ạ, em đang vội mong mn giúp đỡ
