=(7y-3x-34)(7y+3x-22)

[7(y-4)]^2 -[3(y+2)]^2 = [7(y-4)-3(y+2)]*[7(y-4)+3(y+2)]

= (4y-34)(10y-22)=2(2y-17)(5y-11) 

mình nghĩ thế sai thâu nha !

\(a,49.\left(y-4\right)^2-9y^2-36y-36=49\left(y-4\right)^2-9\left(y^2+4y+4\right)\)

\(=49\left(y-4\right)^2-9\left(y+4\right)^2=\left(7y-28\right)^2-\left(3y+12\right)^2\)

\(=\left(7y-28+3y+12\right)\left(7y-28-3y-12\right)\)

\(=\left(10y-16\right)\left(4y-40\right)=8\left(5y-8\right)\left(y-10\right)\)

\(b,xyz-\left(xy+yz+xz\right)+\left(x+y+z\right)-1\)

\(=xyz-xy-yz-xz+x+y+z-1\)

\(=\left(xyz-xy\right)-\left(xz-x\right)-\left(yz-y\right)+\left(z-1\right)\)

\(=xy\left(z-1\right)-x\left(z-1\right)-y\left(z-1\right)+\left(z-1\right)\)

\(=\left(z-1\right)\left(xy-x-y+1\right)\)

\(=\left(z-1\right)\text{[}x\left(y-1\right)-\left(y-1\right)\text{]}\)

\(=\left(z-1\right)\left(y-1\right)\left(x-1\right)\)

x²(x - y) + 49(y - x)

= x²(x - y) - 49(x - y)

= (x² - 49)(x - y)

= (x - 7)(x + 7)(x - y)

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

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

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

`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 : 

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

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

1)\(36\left(x-y\right)^2-49\left(x+y\right)^2\) 

\(=\left(6x-6y+7x+7y\right)\left(6x-6y-7x-7y\right)\)

\(=\left(13x+y\right)\left(-x-13y\right)\)

\(=-\left(13x+y\right)\left(x+13y\right)\)

2)\(16x^2-9\left(x+y\right)^2=\left(4x+3x+3y\right)\left(4x-3x-3y\right)=\left(7x+3y\right)\left(x-3y\right)\)