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

`9(x-y)^2-4(x+y)^2`

`=[3(x-y)]^2-[2(x+y)]^2`

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

`=(3x-3y+2x+2y)(3x-3y-2x-2y)`

`=(5x-y)(x-5y)`

\(9\left(x-y\right)^2-4\left(x+y\right)^2\\ =\left[3\left(x-y\right)\right]^2-\left[2\left(x+y\right)\right]^2\\ =\left[3\left(x-y\right)-2\left(x+y\right)\right]\left[3\left(x-y\right)+2\left(x+y\right)\right]\\ =\left(3x-3y-2x-2y\right)\left(3x-3y+2x+2y\right)\\ =\left(x-5y\right)\left(5x-y\right)\)

Ta có: \(9\left(x-y\right)^2-4\left(x+y\right)^2\)

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

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

\(=\left(x-5y\right)\left(5x-y\right)\)

Bài làm:

Ta có: \(9\left(x-y\right)^2-4\left(x+y\right)^2=\left[3\left(x-y\right)\right]^2-\left[2\left(x+y\right)\right]^2\)

\(=\left(3x-3y-2x-2y\right)\left(3x-3y+2x+2y\right)=\left(x-5y\right)\left(5x-y\right)\)

Học tốt!!!!

Ta có : 

\(9\left(x-y\right)^2-4\left(x+y\right)^2=9x^2-18xy+9y^2-4x^2-8xy-4y^2\)

\(=5x^2-26xy+5y^2==\left(5x-y\right)\left(x-5y\right)\)

\(9(x-y)^2-27(y-x)^3\\=9(x-y)^2+27(x-y)^3\\=(x-y)^2[9+27(x-y)]\\=(x-y)^2(9+27x-27y)\\=9(x-y)^2(1+3x-3y)\)

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

a: 2x+4=2(x+2)

b: \(x^2+2xy+y^2-9=\left(x+y-3\right)\left(x+y+3\right)\)

a) (x-y)²-4=(x-y)²-2²=(x-y-2)(x-y+2)

b) 9-(x-y)²=3²-(x-y)²=(3-x+y)(3+x-y)

c) (x²+4)²-16x²=(x²+4)²-(4x)²=(x²+4-4x)(x²+4+4x)=(x-2)²(x+2)²

a) (x - y)² -  4

= (x - y)² -  2²

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

b) 9 - (x - y)²

= 3² - (x - y)²

= (3 - x + y)(3 + x - y)

c) (x² + 4)² - 16x²

= (x² + 4)² - (4x)²

= (x² + 4 - 4x)(x² + 4 + 4x)

= [(x + 2)(x - 2)]²

= (x² + 4)²

`a,`

`(x-y)^2 -4`

`= (x-y)^2 - 2^2`

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

`b,`

`9 - (x-y)^2`

`=3^2 - (x-y)^2`

`= (3-x+y) (3 +x-y)`

`c,`

`(x^2 +4)^2 -16x^2`

`= (x^2 +4)^2 - (4x)^2`

`= (x^2 +4-4x)(x^2 +4+4x)`

`= (x^2 - 2 . x . 2 +2^2) (x^2 +2 . x . 2 +2^2)`

`= (x-2)^2 (x+2)^2`

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

a: =(x-y)(5-y)

b: \(=x^2-6x+9-y^2=\left(x-3-y\right)\left(x-3+y\right)\)

\(a,5\left(x-y\right)-y\left(x-y\right)=\left(5-y\right)\left(x-y\right)\\ b,x^2-6x-y^2+9=\left(x^2-6x+9\right)-y^2=\left(x-3\right)^2-y^2=\left(x-y-3\right)\left(x+y-3\right)\)