a) 2x² - xy + 4x - 2y
<=> (2x² + 4x)-(xy + 2y)
<=> 2x(x + 2) - y(x + 2)
<=> (x + 2)(2x - y)
b) (a²−a+2012)(a²−a+2014)−3
 Đặt a²−a+2012 là x , ta có : 
  x(x + 2) - 3
<=> x² + 2x - 3
<=> x² + 3x - x - 3
<=> x(x + 3) - (x + 3)
<=> (x +3)(x - 1)
  Thay x = a²−a+2012 , ta được :
    (a²−a+2015)(a²−a+2011)

 

giúp mik vs

ta có: (a^2-a+2012)(a^2-a+2014)-3

=(a^2-a+2013-1)(a^2-a+2013+1)-3

=(a^2-a+2013)^2-1-3

=(a^2-a+2013)^2-4

=(a^2-a+2013-2)(a^2-a+2013+2)

=(a^2-a+2011)(a^2-a+2015)

chúc bn học tốt

= x³ + y³ + z³ + 3x²yz + 3xy²z + 3xyz² - x³ -y³ - z³

=3x²yz + 3xy²z + 3xyz²

= 3xyz( x + y + z)

b.

x^4+2012x^2+2012x-x+2012=

(x^4-x)+2012(x^2+x+1)=

x(x-1)(x^2+x+1)+2012(x^2+x+1)=

(x+2012)(x^2+x+1)

làm sao ra vậy

\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)

\(a^4-a^3-a^2+a\)

\(=a^3\left(a-1\right)-a\left(a-1\right)\)

\(=\left(a-1\right)\left(a^3-a\right)\)

\(\left(a^2-b^2\right)+\left(a^3+b^3\right)-a^2b^2\left(a+b\right)\)

\(=\left(a+b\right)\left(a-b\right)+\left(a+b\right)\left(a^2-ab+b^2\right)-a^2b^2\left(a+b\right)\)

\(=\left(a+b\right)\left(a-b+a^2+b^2-ab-a^2b^2\right)\)

\(=\left(a+b\right)\left[b^2\left(1-a^2\right)+a\left(1+a\right)-b.\left(1+a\right)\right]\)

\(=\left(a+b\right)\left(a+1\right)\left(b^2+a-b\right)\)

\(a^5+a^3-a^2-1=a^3\left(a^2+1\right)-\left(a^2+1\right)=\left(a^3-1\right)\left(a^2+1\right)\)

\(a^5+a^3-a^2-1=a^3\left(a^2+1\right)-\left(a^2+1\right)=\left(a^3-1\right)\left(a^2+1\right)=\left(a-1\right)\left(a^2+a+1\right)\left(a^2+1\right)\)

\(a^5+a^3-a^2-1\)

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

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

trả lời thiếu
 

\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)