=(x-6)(x+1)

a,

C=\(4.x^2-9.y^2\)=\(\left(2y\right)^2-\left(3y\right)^2\)=\(\left(2x-3y\right).\left(2x+3y\right)\)

b,để các đường thẳng y=2x+m và y=x-2m+3 cắt nhau tại một điểm nằm trên trục tung thì m=-2m+3

<=>m=1

a)C=4x²-9y²=(2x-3y)(2x+3y)

b) hai đường thẳng y=2x+m và y=x-2m+3 cắt nhau tại 1 điểm trên trục tung khi 

2 khác 1 (thỏa mãn) và m=-2m+3 <=> 3m=3<=> m=1

x² - 4x - y² + 4

= (x² - 4x + 4) - y²

= (x - 2)² - y²

= (x - y - 2)(x + y - 2)

(x - y - 2)(x + y - 2)

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

d) \(\left(x-1\right)^2-4\left(2x-3\right)^2=\left(5-3x\right)\left(5x-7\right)\)

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

e) \(x^2-7=\left(x-\sqrt{7}\right)\left(x+\sqrt{7}\right)\)

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

d) \(\left(x-1\right)^2-4\left(2x-3\right)^2=\left(5x-7\right)\left(5-3x\right)\)

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

f) \(x^2-7=\left(x-\sqrt{7}\right)\left(x+\sqrt{7}\right)\)

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

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

\(1,=x\left(x+y\right)+\left(x+y\right)=\left(x+1\right)\left(x+y\right)\\ 2,\Leftrightarrow x^2-3x-4x+12=0\\ \Leftrightarrow\left(x-3\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

=(x²+x)+(xy+y)

=x(x+1)+y(x+1)

=(x+y)(x+1)

a) 2x²-6x-x+3 = 2x(x-3) - (x-3) = (x-3)(2x-1)

b) x²-x-5x+5 = x(x-1) - 5(x-1) = (x-1)(x-5)

c) 5x(x-2y) + 2( x-2y)² = (x-2y)(5x+2x-2y) = (x-2y)(7x-2y)

chú ý : (A-B)²=(B-A)²

d) 7x(4-y)² - (4-y)³ = ( 16-8y+y²) (7x-4+y)

a) \(2x^2-7x+3=2x^2-6x-x+3=2x\left(x-3\right)-\left(x-3\right)=\left(x-3\right)\left(2x-1\right)\)

b) \(x^2-6x+5=x^2-5x-x+5=x\left(x-5\right)-\left(x-5\right)=\left(x-5\right)\left(x-1\right)\)

c)\(5x\left(x-2y\right)+2\left(2y-x\right)^2=5x\left(x-2y\right)+2\left(x-2y\right)^2\\ =\left(x-2y\right)\left(5x+2x-4y\right)=\left(x-2y\right)\left(7x-4y\right)\)

d) \(7x\left(y-4\right)^2-\left(4-y\right)^3=7x\left(y-4\right)+\left(y-4\right)^3=\left(y-4\right)\left(7x-y-4\right)\)